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

Last Updated : 12 Sep, 2022

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

Input : N = 5
Output : 55```

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

 `// 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`

Time Complexity : O(n)

Auxiliary Space: O(1)

Method 2:

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 * ? k2 + 3 * ? k + ? 1
n3 + 3 * n2 + 3 * n + 1 = 1 + 3 * ? k2 + 3 * (n * (n + 1))/2 + n
n3 + 3 * n2 + 3 * n = 3 * ? k2 + 3 * (n * (n + 1))/2 + n
n3 + 3 * n2 + 2 * n - 3 * (n * (n + 1))/2 = 3 * ? k2
n * (n2 + 3 * n + 2) - 3 * (n * (n + 1))/2 = 3 * ? k2
n * (n + 1) * (n + 2) - 3 * (n * (n + 1))/2 = 3 * ? k2
n * (n + 1) * (n + 2 - 3/2) = 3 * ? k2
n * (n + 1) * (2 * n + 1)/2  = 3 * ? k2
n * (n + 1) * (2 * n + 1)/6  = ? k2```

## Java

 `// 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 is 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

 `// 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`

Time complexity: O(1) as it is doing constant operations

Auxiliary Space: O(1) as it is using constant space

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

Article Tags :