# C++ Program for cube sum of first n natural numbers

Print the sum of series 1^{3} + 2^{3} + 3^{3} + 4^{3} + …….+ n^{3} till n-th term.

Examples:

Input : n = 5 Output : 225 1^{3}+ 2^{3}+ 3^{3}+ 4^{3}+ 5^{3}= 225 Input : n = 7 Output : 784 1^{3}+ 2^{3}+ 3^{3}+ 4^{3}+ 5^{3}+ 6^{3}+ 7^{3}= 784

## CPP

## CPP

`// Simple C++ program to find sum of series` `// with cubes of first n natural numbers` `#include <iostream>` `using` `namespace` `std;` `/* Returns the sum of series */` `int` `sumOfSeries(` `int` `n)` `{` ` ` `int` `sum = 0;` ` ` `for` `(` `int` `x=1; x<=n; x++)` ` ` `sum += x*x*x;` ` ` `return` `sum;` `}` `// Driver Function` `int` `main()` `{` ` ` `int` `n = 5;` ` ` `cout << sumOfSeries(n);` ` ` `return` `0;` `}` |

**Output :**

225

Time Complexity : O(n)

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

**Output: **

225

Time Complexity : O(1) **How does this formula work?**

We can prove the formula using mathematical induction. We can easily see that the formula holds true for n = 1 and n = 2. Let this be true for n = k-1.

Let the formula be true for n = k-1. Sum of first (k-1) natural numbers = [((k - 1) * k)/2]^{2}Sum of first k natural numbers = = Sum of (k-1) numbers + k^{3}= [((k - 1) * k)/2]^{2}+ k^{3}= [k^{2}(k^{2}- 2k + 1) + 4k^{3}]/4 = [k^{4}+ 2k^{3}+ k^{2}]/4 = k^{2}(k^{2}+ 2k + 1)/4 = [k*(k+1)/2]^{2}

**The above program causes overflow, even if result is not beyond integer limit.** Like previous post, we can avoid overflow upto some extent by doing division first.

## CPP

## CPP

`// Efficient CPP program to find sum of cubes` `// of first n natural numbers that avoids` `// overflow if result is going to be within` `// limits.` `#include<iostream>` `using` `namespace` `std;` ` ` `// Returns sum of first n natural` `// numbers` `int` `sumOfSeries(` `int` `n)` `{` ` ` `int` `x;` ` ` `if` `(n % 2 == 0)` ` ` `x = (n/2) * (n+1);` ` ` `else` ` ` `x = ((n + 1) / 2) * n;` ` ` `return` `x * x;` `}` ` ` `// Driver code` `int` `main()` `{` ` ` `int` `n = 5;` ` ` `cout << sumOfSeries(n);` ` ` `return` `0;` `}` |

**Output: **

225

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