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

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
```
 `// 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.

 `// 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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