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C++ Program for cube sum of first n natural numbers
• Last Updated : 09 Oct, 2018

Print the sum of series 13 + 23 + 33 + 43 + …….+ n3 till n-th term.

Examples:

```Input : n = 5
Output : 225
13 + 23 + 33 + 43 + 53 = 225

Input : n = 7
Output : 784
13 + 23 + 33 + 43 + 53 +
63 + 73 = 784
```

## CPP

 `// Simple C++ program to find sum of series``// with cubes of first n natural numbers``#include ``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 + k3
= [((k - 1) * k)/2]2 + k3
= [k2(k2 - 2k + 1) + 4k3]/4
= [k4 + 2k3 + k2]/4
= k2(k2 + 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

 `// Efficient CPP program to find sum of cubes ``// of first n natural numbers that avoids ``// overflow if result is going to be withing ``// limits.``#include``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
```

Please refer complete article on Program for cube sum of first n natural numbers for more details!

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