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The correct option is C. 1296. The **sum** **of** the **cubes** **of** first **n** **natural** **numbers** is given by n2(n+1)24. Substituting n=8 , we get **sum** as 1296. Suggest Corrections. Then, write the value of. ∑ r = 1 **n** S r s r. Q. Let S **n** denote the **sum** of the **cubes** of the first **n natural numbers** and S **n** denote the **sum** of the first **natural numbers**. Then ∑ r = 1 **n** S r S r equals. (a) **n n** + 1 **n** + 2 6. (b) **n n** + 1 2. (c) **n** 2 + 3 **n** + 2 2. (d) None of these.

## tc

The **sum** of the squares of the first **n** odd **natural numbers** is given by 1 2 + 3 2 + 5 2 + + (2n - 1) 2.. When odd **numbers** are cubed , the result will be an odd **number** . For example, 3 3 = 27, 5 3 = 125; Perfect **cube numbers** can be both positive or negative integers. For example, both 8 and - 8 are the perfect **cubes** of 2 and - 2 respectively. Program to find the **sum** **of** first **n** **natural** **numbers**. We will see two C programs to calculate the **sum** **of** **natural** **numbers**. In the first C program we are using for loop for find the **sum** and in the second program we are doing the same using while loop. To understand these programs,. **Sum** **of** **cubes** **of** first **n** **natural** **numbers** is given by below formula \[\sum_{1}^{n} n^{3}=(\frac{n(n+1)}{2})^{2} \] where **n** is **natural** **number**. This method can easily be employed to calculate the **cubes** **of** **n** **natural** **numbers** using the above formula. For example \[\sum_{1}^{5} n^{3}= 1^{3}+2^{3}+3^{3}+4^{3}+5^{3} \]. The **sum** of the squares of the first **n** odd **natural numbers** is given by 1 2 + 3 2 + 5 2 + + (2n - 1) 2.. When odd **numbers** are cubed , the result will be an odd **number** . For example, 3 3 = 27, 5 3 = 125; Perfect **cube numbers** can be both positive or negative integers. For example, both 8 and - 8 are the perfect **cubes** of 2 and - 2 respectively.. Find the **number** **of** ways that a given integer, , can be expressed as the **sum** **of** the powers of unique, **natural** **numbers**. For example, if and , we have to find all combinations of unique squares adding up to . The only solution is . Function Description Complete the powerSum function in the editor below. Explanation for correct option. The first term of the **natural number** is a 1 = 1. The second term of the **natural number** is a 2 = 2. The common difference is d = 1. Total **number** of terms = **n**. We. **Sum** **of** **cubes** **of** first **n** **natural** **numbers**: We determine the **sum** **of** **cubes** **of** consecutive **natural** **numbers** by the following formula: Prove that: 1 3 + 2 3 + 3 3 + ⋯ + **n** 3 = [ **n** ( **n** + 1) 2] 2. Proof: Let S denote the desired **sum**. So we have S = 1 3 + 2 3 + ⋯ + **n** 3. To prove the formula, we will use the fact below: **n** 4 − ( **n** − 1) 4 = 4 **n** 3 −.

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## ww

The **sum** of the squares of the first **n** odd **natural numbers** is given by 1 2 + 3 2 + 5 2 + + (2n - 1) 2.. When odd **numbers** are cubed , the result will be an odd **number** . For example, 3 3 = 27, 5 3 = 125; Perfect **cube numbers** can be both positive or negative integers. For example, both 8 and - 8 are the perfect **cubes** of 2 and - 2 respectively. 2017. 2. 28. · This mimics our development of the **natural numbers**. It is also equivalent to prove that whenever the conjecture is true for **n** − 1 , {\displaystyle **n**-1,} it's true for **n** . {\displaystyle. Every **number** can be represented by the **sum** **of** nine **cubes**. Indeed every **number** but 23 = 2·23 +7·13 and 239 = 2·43 +4·33 +3·13 can be expressed by the **sum** **of** at most eight **cubes**. Moreover, there are only 15 integers that require 8 **cubes**. This implies that G(3) ≤ 7. G(k) is only known for the cases k = 2,4. The best currently known bound. In other words, the **sum** **of** the first **n** **natural** **numbers** is the **sum** **of** the first **n** **cubes**. Formula to Find **Sum** **of** **Cubes** The other name for the formula of **sum** **of** **cube** is factoring formula. The find the **sum** **of** **cubes** **of** any polynomial the given formula is used: a 3 + b 3 = (a + b) (a 2 − ab + b 2) Solved Example Question Question:Factor 27 x3 + 1.

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## th

**Sum** **of** **cubes** **of** first **n** **natural** **numbers** is . ... **Sum** **of** fourth powers of **n** **natural** **numbers** is **Sum** **of** fifth powers of **n** **natural** **numbers** is . **Sum** **of** sixth powers of **n** **natural** **numbers** is . It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel. Sum of cubes of first n natural numbers can be calculated using the formula [n* (n+1)/2]^2. Sum of cubes of first 4 natural numbers (1 ->1, 2 ->8, 3 ->27, 4->64) is 100 and is represented as Sn =. **Sum** **of** First **n** Odd **Natural** **Numbers**. 2.1.5. Pythagorean Triplets. 2.1.6. Using Identities to Find the Square of a **Number**. 2.1.7. Product of Two Consecutive Odd or Consecutive Even **Numbers**. 2.2. Square Root. 2.3. Square Root by Long-Division Method 3. **Cube** and **Cube** Roots. 4..

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## uz

In order to find **sum** we need to iterate through all **natural numbers** between 1 to **n**. Initialize a loop from 1 to **N**, increment loop counter by 1 for each iteration. C Program to Find **Sum** of **Natural Numbers** Using While Loop. Aug 09, 2010 · Write a program to input a **number**. Find the **sum** of digits and the **number** of digits. Display the output. The smallest **natural number** is 1. Objective: Write a Python program which returns **sum** of **cubes** of **natural numbers** starting from 1 to given **natural number n**, (1 3 + 2 3 + 3 3 + ... + **n** 3).. The **sum** of the squares of the first **n** odd **natural numbers** is given by 1 2 + 3 2 + 5 2 + + (2n - 1) 2.. When odd **numbers** are cubed , the result will be an odd **number** . For example, 3 3 = 27, 5 3 = 125; Perfect **cube numbers** can be both positive or negative integers. For example, both 8 and - 8 are the perfect **cubes** of 2 and - 2 respectively.. Enter an positive integer: 10 **Sum** = 55 In this program, the **number** entered by the user is passed to the add () function. Suppose, 10 is entered by the user. Now, 10 is passed to the add () function. This function adds 10 to the addition result of 9 (10 - 1 = 9). Next time, 9 is added to the addition result of 8 (9 - 1 = 8). 1 day ago · The **sum** of the **cubes** of the first **n natural numbers** is 2025, then find the value of **n** . Tamil Nadu Board of Secondary Education SSLC (English Medium) Class 10th ... The **sum** of. In the mathematics of **sums** **of** powers, it is an open problem to characterize the **numbers** that can be expressed as a **sum** **of** three **cubes** **of** integers, allowing both positive and negative **cubes** in the **sum**. A necessary condition for to equal such a **sum** is that cannot equal 4 or 5 modulo 9, because the **cubes** modulo 9 are 0, 1, and −1, and no three of these **numbers** can **sum** to 4 or 5 modulo 9.

Question of Class 8-exercise-4-true-and-false- : The **sum** **of** the **cubes** **of** first **n** **natural** **numbers** is equal to the squares of their **sum** . ... The **sum** **of** the **cubes** **of** first **n** **natural** **numbers** is equal to the squares of their **sum** . Talk to Our counsellor: Give a missed call 07019243492. Login / Register. Notes.

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## my

Click here👆to get an answer to your question ️ Find the **sum** of the **cubes** of first five **natural numbers**. Verify your answer by using formula (**n**)(**n** + 1)/2]^2 , where **n** is a **natural numbers**. Also find the **sum** of the cubs of the first ten **natural numbers**. 1. Write definitions for the following two functions using Python: (30 pts) sumN (**n**) returns the **sum** **of** the first **n** **natural** **numbers**. sumNCubes (**n**) returns the **sum** **of** the **cubes** **of** the first **n** **natural** **numbers**. Then use these functions in a program that prompts a user to enter a **number** **'n'** and print out the **sum** **of** the first **n** **natural** **numbers**, and.

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## te

This item: Yogasleep Rohm Portable White Noise Machine + Travel Case 3 Soothing, **Natural** Sounds with Volume Control Compact Sleep Therapy USB Rechargeable 3-Pack, Rohm + Travel Case 3-Pack Bundle $120.00 ... such as finding the **sum** of digits of all **numbers** preceding each **number** in a series.

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## kw

Answer - We know that there will be 2n **numbers** in between the squares of the **numbers n** and (**n** + 1). (i) Between 12 2 and 13 2, there will be 2 × 12 = 24 **numbers** (ii) Between 25 2 and 26 2, ... To find the square root and **cube** root of large **numbers**, we have to use prime factorization method. An example is shown below: Find the square root of 144. Formula to find the sum of cubes first n natural numbers : 1 3 + 2 3 + 3 3 + ........ + n 3 = [n (n + 1)/2]2. Proof : To find ( 12 + 22 + 32 + ........ + n2 ), let us consider the identity. (x + 1) 4 - x 4 = 4x. Mathematical Formula. Following is the representation to find the **sum** **of** **n** **natural** **numbers** using the mathematical formula: **Sum** **of** **n** **natural** **number** = **n** * (**n** + 1) / 2. Where **n** defines the **natural** **number**. Suppose, we want to calculate the **sum** **of** the first 20 **natural** **number**, we need to put in a mathematical formula to get the **sum**:.

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## sf

Proof by induction Introduction. In FP1 you are introduced to the idea of proving mathematical statements by using induction.. Proving a statement by induction follows this logical structure.

**Sum** **of** squares of first **N** **natural** **numbers** is given as = 12 + 22 + 32 + 42 + ...... + (n-1)2 + n2 To find the **sum** **of** the squares of **N** **natural** **numbers** declare a variable and initialize it with value 1. Declare another variable **sum** and initialize it with 0. Now, in every iteration increase **sum** value until the value **N**. The base condition for recursion is defined and if the input **number** is less than or equals to 1, the **number** is returned, else we return the same function call with **number** decremented by 1. In this way, the recursive function works in Python that can calculate the **sum** **of** **natural** **numbers**.

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## rd

T(**n**) = **n**(n+1)/2, where T(**n**) represents the **sum** of the first **n natural numbers**.For our fifth and final look at deriving the closed form formula for the **sum** of the first **n natural numbers**, we will start by taking a look at the **sum** of the first **n** integers squared (from 0 to **n**) as well as the first **n** integers squared (from 0 to n+1). How to Write PseudoCode to Find **Sum** of **Natural Numbers**.

Then, write the value of. ∑ r = 1 **n** S r s r. Q. Let S **n** denote the **sum** of the **cubes** of the first **n natural numbers** and S **n** denote the **sum** of the first **natural numbers**. Then ∑ r = 1 **n** S r S r equals. (a) **n n** + 1 **n** + 2 6. (b) **n n** + 1 2. (c) **n** 2 + 3 **n** + 2 2. (d) None of these.

To find the **sum** **of** **cubes** **of** the first **n** **natural** **numbers**, the code is as follows − Example Live Demo <?php function sum_of_cubes($val) { $init_sum = 0; for ($x = 1; $x <= $val; $x++) $init_sum += $x * $x * $x; return $init_sum; } print_r("The **sum** **of** **cubes** **of** first 8 **natural** **numbers** is "); echo sum_of_cubes(8); ?> Output.

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## pl

2022. 9. 10. · In this **cube sum** program , we find **cube** of individual digit of a **number** and find their **sum** .for example: 123 = 1^3 + 2 ^3 + 3^3. **Cube** of individual digit have to be found for that we.

Write a recursive function that calculate **sum** **of** first **n** **natural** **numbers**.

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## hb

The formula to find the **sum** **of** **cubes** **of** **n** **natural** **numbers** is S = [**n** 2 (**n** + 1) 2 ]/4, where **n** is the count of **natural** **numbers** that we take. For example, if you want to find the **sum** **of** **cubes** **of** 7 **natural** **numbers**, you will put the value of **n** as 7 in the formula. What is the **Sum** **of** **Cubes** **of** First 20 **Natural** **Numbers**?. In order to find **sum** we need to iterate through all **natural numbers** between 1 to **n**. Initialize a loop from 1 to **N**, increment loop counter by 1 for each iteration. C Program to Find **Sum** of **Natural Numbers** Using While Loop. Aug 09, 2010 · Write a program to input a **number**. Find the **sum** of digits and the **number** of digits. Display the output.

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## cj

Given a **number** s (1 <= s <= 1000000000). If s is **sum** **of** the **cubes** **of** the first **n** **natural** **numbers** then print **n**, otherwise print -1. First few Squared triangular **number** are 1, 9, 36, 100, 225, 441, 784, 1296, 2025, 3025, Examples : Input : 9 Output : 2 Explanation : The given **number** is **sum** **of** **cubes** **of** first 2 **natural** **numbers**. **Number** **of** terms given **sum** **of** first **n** terms, first term and last term; **Number** **of** terms of in arithmetic progression; Position of pth term given pth term, first term and common difference; ... **Sum** **of** **cubes** **of** first **n** **natural** **numbers**; **Sum** **of** first **n** even **natural** **numbers**; **Sum** **of** first **n** **natural** **numbers**;. a6_04: **Sum** **Numbers**. Name of files: "sumNumbers.py". Save source file in folder: "a6_04_ LastName ". INSTRUCTIONS: Write definitions for the following two functions: sumN (**n**) returns the **sum** **of** the first **n** **natural** **numbers**. sumNCubes (**n**) returns the **sum** **of** the **cubes** **of** the first **n** **natural** **numbers**.

2022. 6. 12. · Method: Finding cube sum of first n natural numbers using built-in function pow (). The pow () function finds the cube of a number by giving the values of i and number. ex: pow.

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## iz

If we may assume to know that **n** ∑ k = 1k = n(n + 1) 2 we can use the partial summation formula **n** ∑ k = 1k2 = **n** ∑ k = 1k ⋅ k = **n** **n** ∑ k = 1k − **n** − 1 ∑ k = 1(k(k + 1) 2)(k + 1 − k) = n2(n + 1) 2 − 1 2n − 1 ∑ k = 1k2 − n(n − 1) 4 so 3 2 **n** ∑ k = 1k2 = 2n2(n + 1) − n(n + 1) + 2n2 4 ⇒ **n** ∑ k = 1k2 = n(n + 1)(2n + 1) 6. answered May 31, 2016 at 14:03.

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## jn

2022. 9. 7. · In the mathematics of **sums** of powers, it is an open problem to characterize the **numbers** that can be expressed as a **sum** of three **cubes** of integers, allowing both positive and. Ask a Question. **Sum** of **cubes** of first **n natural numbers** is _____ ← Prev Question Next Question →.

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## kv

m = (a 3 + b 3) = (c 3 + d 3) for distinct a, b, c, d. For example, If **n** = 25000, m can be any of 1729, 4104, 13832, or 20683 as these **numbers** can be represented as the **sum** **of** two **cubes** for two different pairs. 1729 = 1 3 + 12 3 = 9 3 + 10 3 4104 = 2 3 + 16 3 = 9 3 + 15 3 13832 = 2 3 + 24 3 = 18 3 + 20 3 20683 = 10 3 + 27 3 = 19 3 + 24 3. If the **sum** of the first **n natural numbers** is S 1 and that of their squares is S 2 and **cubes** is S 3 , then show that 9 S 2 2 = S 3 (1 + 8 S 1 ). Medium View solution. Summing **cubes** of **natural numbers** within a range in JavaScript. Jul 20, 2017 · We use = to assign a value on the right to a variable on the left. let age = 27; ... This **sum** variable will hold the total **sum** of all **numbers** and it will be initialized as 0. Javascript program : The javascript program to find.

**Sum** of **cube** of first or consecutive ” **n**” odd **natural numbers** = **n** 2 (2n 2 – 1) Examples on **sum** of **numbers** Ex . 1 : Find the **sum** of the first 50 positive integers. 123 **n** + +++ L. For example, when **n** = , this **sum** is equal to the **number** **of** **cubes** in the figure. There are 55 × **cubes** on the lowest level of the figure, 4 4 × on the level above, and so on until you.

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## ep

The theoretical **sum** would be the same. If you are using floating point then the result could differ. The order of operations of built-in functions like harmonic() is not specified. For doing it yourself, replace 1:n by n:-1:1. Answer - We know that there will be 2n **numbers** in between the squares of the **numbers n** and (**n** + 1). (i) Between 12 2 and 13 2, there will be 2 × 12 = 24 **numbers** (ii) Between 25 2 and 26 2, ... To find the square root and **cube** root of large **numbers**, we have to use prime factorization method. An example is shown below: Find the square root of 144. **Sum of cube** of first **n natural number** is determine by using the formula which is shown below, ∑ **n** 3 = **n** 2 ( **n** + 1) 2 4. For first 15 **natural numbers**, we put **n** = 15 in above equation, ∑ **n** 3 = 15 2 ( 15 + 1) 2 4 = 15 2 × 16 2 4 = 14400. Mean **of cubes** of the first 15 **natural numbers** can be calculated by dividing the **sum** of first 15 **natural**. Here are some methods to solve the above mentioned. Sum_of_Natural_number. Exampe Number=10 Enter **Natural** **Number**:- 10 **natural** **number** is: 1 **natural** **number** is: 2 **natural** **number** is: 3 **natural** **number** is: 4 **natural** **number** is: 5 **natural** **number** is: 6 **natural** **number** is: 7 **natural** **number** is: 8 **natural** **number** is: 9 **natural** **number** is: 10 **Sum** **of** **Natural**.

An efficient solution is to use direct mathematical formula which is (**n** ( **n** + 1 ) / 2) ^ 2, For **n** = 5 **sum** by formula is (5* (5 + 1 ) / 2)) ^ 2 = (5*6/2) ^ 2 = (15) ^ 2 = 225 For **n** = 7, **sum** by formula is (7* (7 + 1 ) / 2)) ^ 2 = (7*8/2) ^ 2 = (28) ^ 2 = 784, C++, Java, Python3, C#, PHP, Javascript, #include <iostream>, using namespace std;. **Sum** **of** squares of first **n** **natural** **numbers**; **Sum** **of** squares of first **n** **natural** **numbers** in constant time; Juggler Sequence; Find all **numbers** having digit product equal to k in 1 to **N**; Find perfect **numbers** in a given range; Multiply integer **number** with 3.5; **Sum** **of** **cubes** **of** **n** even **natural** **numbers**; Average of **cubes** **of** first **n** **natural** **numbers**.

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## fq

Find the **number** **of** ways that a given integer, , can be expressed as the **sum** **of** the powers of unique, **natural** **numbers**. For example, if and , we have to find all combinations of unique squares adding up to . The only solution is . Function Description Complete the powerSum function in the editor below. 3 × 3 -3 = 6 ( 6 is the third **number**) **n** will be 6 × 3 - 3 = 18-3 = 15 Hence, **n** is 15. Question: Consider the following statements: (1) The **sum** **of** **cubes** **of** first 2o **natural** **numbers** is 44400 (2) The **sum** **of** squares of first 20 **natural** **numbers** is 2870. Which of the above statements is/are correct? Only 1st; Only 2nd; Solution: B is the correct. Python-Defining Functions. Write definitions for the following two functions: sumN (**n**) returns the **sum** **of** the first **n** **natural** **numbers**. sumNCubes (**n**) returns the **sum** **of** the **cubes** **of** the first **n** **natural** **numbers**. Then use these functions in a program that prompts a user for an **n** and prints out the **sum** **of** the first **n** **natural** **numbers** and the **sum** **of**.

The **sum** **of** **cube** **of** first 7 **natural** **numbers** is 784 A class named Demo defines a static function that takes a value as parameter. Here, an initial **sum** is defined as 0. Next, a 'for' loop is run over values 1 to the value passed as parameter. This is the value up to which the **cubes** **of** **numbers** beginning from 1 need to be computed.

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2022. 9. 7. · In the mathematics of **sums** of powers, it is an open problem to characterize the **numbers** that can be expressed as a **sum** of three **cubes** of integers, allowing both positive and.

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## yc

Given a **number** s (1 <= s <= 1000000000). If s is **sum** **of** the **cubes** **of** the first **n** **natural** **numbers** then print **n**, otherwise print -1. First few Squared triangular **number** are 1, 9, 36, 100, 225, 441, 784, 1296, 2025, 3025, Examples : Input : 9 Output : 2 Explanation : The given **number** is **sum** **of** **cubes** **of** first 2 **natural** **numbers**. The **sum** **of** the **cubes** **of** the first **n** **natural** **numbers** is 2025, then find the value of **n** . Tamil Nadu Board of Secondary Education SSLC (English Medium) Class 10th ... The **sum** **of** the **cubes** **of** the first **n** **natural** **numbers** is 2025, then find the value of **n**. Advertisement Remove all ads. Solution Show Solution. 1 2 + 2 2 + 3 2 +. 2021. 4. 3. · Here, we will write a Python program to find the **sum** of **cube** of first **N natural numbers**. Submitted by Shivang Yadav, on April 03, 2021 . Python programming language is a.

The theoretical **sum** would be the same. If you are using floating point then the result could differ. The order of operations of built-in functions like harmonic() is not specified. For doing it yourself, replace 1:n by n:-1:1.

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## sr

Calculate **n** terms of square **natural** **number** and their **sum**: ----- Input the **number** **of** terms : 10 The square **natural** upto 10 terms are :1 4 9 16 25 36 49 64 81 100 The **Sum** **of** Square **Natural** **Number** upto 10 terms = 385.

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## mz

"Every **natural** **number** is either a **cube** **number** or the **sum** **of** 2,3,4,5,6,7,8 or 9 **cube** **numbers**." (2), page 37ff. That means that 9 is a smallest **number**. It can be more than 9 as the following **sum** **of** 180³ with 64 (!) cubic **numbers** shows. 180³ = 6³+7³+8³+...+67³+68³+69³ (1). Already 4 summands will do, 180=1³+3³+3³+5³. The first **numbers** 1=1³ 2=1³+1³. Proof by induction Introduction. In FP1 you are introduced to the idea of proving mathematical statements by using induction.. Proving a statement by induction follows this logical structure.

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## kf

On substituting the **n** value in formula and further simplify by using a basic arithmetic operation to get the required solution. Complete answer: Before solving the problem, we will discuss the formula of **sum** **of** the **cubes** **of** first **n** **natural** **numbers**. The series. S = ∑ k = 1 **n** K 3 = 1 3 + 2 3 + 3 3 +... + **n** 3. gives the **sum** **of** the **cube** **of**.

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## cx

This is a visual proof for why the **sum** **of** first **n** **cubes** is the square of the **sum** **of** first **n** **natural** **numbers**. Traditionally, it is proved algebraically using binomial theorem, **sum** **of** squares formula and the **sum** **of** **natural** **numbers**, but this is a very elegant proof from Nelsen - Proof without words. More visual proofs from the same book on the. Algorithm: Take input of **n** till which we need to get the **sum**. Initialize a variable **sum** and declare it equal to 0 (to remove garbage values). Using while loop, add all **numbers** 1 to **n**. Now, Print the **sum**. n(n−1)(n+1) and 3n are both divisible by 3. Hence their **sum** is also divisible by 3.

2017. 11. 7. · Sum of cube of first or consecutive ” n” natural numbers: Sum of cube of first or consecutive ” n” even natural numbers = 2n2 (n + 1)2 Sum of cube of first or consecutive ” n” odd natural numbers = n2 (2n2 – 1) Examples on. Output. Enter a **number**: 10 [1] "The **sum** is 55". Here, we ask the user for a **number** and display the **sum** **of** **natural** **numbers** upto that **number**. We use while loop to iterate until the **number** becomes zero. On each iteration, we add the **number** num to **sum**, which gives the total **sum** in the end. We could have solved the above problem without using any.

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