# Java Program for Sum of squares of first n natural numbers

Given a positive integer N. The task is to find 12 + 22 + 32 + ….. + N2.

Examples:

```Input : N = 4
Output : 30
12 + 22 + 32 + 42
= 1 + 4 + 9 + 16
= 30

Iput : N = 5
Output : 55
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1: O(N) The idea is to run a loop from 1 to n and for each i, 1 <= i <= n, find i2 to sum.

 `// Java Program to find sum of ` `// square of first n natural numbers ` `import` `java.io.*; ` ` `  `class` `GFG { ` ` `  `    ``// Return the sum of square of first n natural numbers ` `    ``static` `int` `squaresum(``int` `n) ` `    ``{ ` `        ``// Iterate i from 1 and n ` `        ``// finding square of i and add to sum. ` `        ``int` `sum = ``0``; ` `        ``for` `(``int` `i = ``1``; i <= n; i++) ` `            ``sum += (i * i); ` `        ``return` `sum; ` `    ``} ` ` `  `    ``// Driven Program ` `    ``public` `static` `void` `main(String args[]) ``throws` `IOException ` `    ``{ ` `        ``int` `n = ``4``; ` `        ``System.out.println(squaresum(n)); ` `    ``} ` `} ` ` `  `/*This code is contributed by Nikita Tiwari.*/`

Output:

```30
```

Method 2: O(1) Proof:

```We know,
(k + 1)3 = k3 + 3 * k2 + 3 * k + 1
We can write the above identity for k from 1 to n:
23 = 13 + 3 * 12 + 3 * 1 + 1 ......... (1)
33 = 23 + 3 * 22 + 3 * 2 + 1 ......... (2)
43 = 33 + 3 * 32 + 3 * 3 + 1 ......... (3)
53 = 43 + 3 * 42 + 3 * 4 + 1 ......... (4)
...
n3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 ......... (n - 1)
(n + 1)3 = n3 + 3 * n2 + 3 * n + 1 ......... (n)

Putting equation (n - 1) in equation n,
(n + 1)3 = (n - 1)3 + 3 * (n - 1)2 + 3 * (n - 1) + 1 + 3 * n2 + 3 * n + 1
= (n - 1)3 + 3 * (n2 + (n - 1)2) + 3 * ( n + (n - 1) ) + 1 + 1

By putting all equation, we get
(n + 1)3 = 13 + 3 * Σ k2 + 3 * Σ k + Σ 1
n3 + 3 * n2 + 3 * n + 1 = 1 + 3 * Σ k2 + 3 * (n * (n + 1))/2 + n
n3 + 3 * n2 + 3 * n = 3 * Σ k2 + 3 * (n * (n + 1))/2 + n
n3 + 3 * n2 + 2 * n - 3 * (n * (n + 1))/2 = 3 * Σ k2
n * (n2 + 3 * n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2
n * (n + 1) * (n + 2) - 3 * (n * (n + 1))/2 = 3 * Σ k2
n * (n + 1) * (n + 2 - 3/2) = 3 * Σ k2
n * (n + 1) * (2 * n + 1)/2  = 3 * Σ k2
n * (n + 1) * (2 * n + 1)/6  = Σ k2
```

 `// Java Program to find sum ` `// of square of first n ` `// natural numbers ` `import` `java.io.*; ` ` `  `class` `GFG { ` ` `  `    ``// Return the sum of square ` `    ``// of first n natural numbers ` `    ``static` `int` `squaresum(``int` `n) ` `    ``{ ` `        ``return` `(n * (n + ``1``) * (``2` `* n + ``1``)) / ``6``; ` `    ``} ` ` `  `    ``// Driven Program ` `    ``public` `static` `void` `main(String args[]) ` `        ``throws` `IOException ` `    ``{ ` `        ``int` `n = ``4``; ` `        ``System.out.println(squaresum(n)); ` `    ``} ` `} ` ` `  `/*This code si contributed by Nikita Tiwari.*/`

Output:

```30
```

Avoiding early overflow:
For large n, the value of (n * (n + 1) * (2 * n + 1)) would overflow. We can avoid overflow up to some extent using the fact that n*(n+1) must be divisible by 2.

 `// Java Program to find sum of square of first ` `// n natural numbers. This program avoids ` `// overflow upto some extent for large value ` `// of n. ` ` `  `import` `java.io.*; ` `import` `java.util.*; ` ` `  `class` `GFG { ` `    ``// Return the sum of square of first n natural ` `    ``// numbers ` `    ``public` `static` `int` `squaresum(``int` `n) ` `    ``{ ` `        ``return` `(n * (n + ``1``) / ``2``) * (``2` `* n + ``1``) / ``3``; ` `    ``} ` ` `  `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `n = ``4``; ` `        ``System.out.println(squaresum(n)); ` `    ``} ` `} ` ` `  `// Code Contributed by Mohit Gupta_OMG <(0_o)> `

Output:

```30
```

Please refer complete article on Sum of squares of first n natural numbers for more details!

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