# Sum of squares of first n natural numbers

• Difficulty Level : Easy
• Last Updated : 15 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.

Below is the implementation of this approach

## C++

 `// CPP Program to find sum of square of first n natural numbers``#include ``using` `namespace` `std;` `// Return the sum of square of first n natural numbers``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``int` `main()``{``    ``int` `n = 4;``    ``cout << squaresum(n) << endl;``    ``return` `0;``}`

## 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.*/`

## Python3

 `# Python3 Program to``# find sum of square``# of first n natural``# numbers`  `# Return the sum of``# square of first n``# natural numbers``def` `squaresum(n) :` `    ``# Iterate i from 1``    ``# and n finding``    ``# square of i and``    ``# add to sum.``    ``sm ``=` `0``    ``for` `i ``in` `range``(``1``, n``+``1``) :``        ``sm ``=` `sm ``+` `(i ``*` `i)``    ` `    ``return` `sm` `# Driven Program``n ``=` `4``print``(squaresum(n))` `# This code is contributed by Nikita Tiwari.*/`

## C#

 `// C# Program to find sum of``// square of first n natural numbers``using` `System;` `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()``    ``{``        ``int` `n = 4;``        ` `        ``Console.WriteLine(squaresum(n));``    ``}``}` `/* This code is contributed by vt_m.*/`

## PHP

 ``

## Javascript

 ``

Output :

`30`

Time Complexity: O(n)

Auxiliary Space: O(1)

Method 2: O(1)

Sum of squares of first N natural numbers = (N*(N+1)*(2*N+1))/6

For example
For N=4, Sum = ( 4 * ( 4 + 1 ) * ( 2 * 4 + 1 ) ) / 6
= 180 / 6
= 30
For N=5, Sum = ( 5 * ( 5 + 1 ) * ( 2 * 5 + 1 ) ) / 6
= 55

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```

Below is the implementation of this approach:

## C++

 `// CPP Program to find sum``// of square of first n``// natural numbers``#include ``using` `namespace` `std;` `// Return the sum of square of``// first n natural numbers``int` `squaresum(``int` `n)``{``    ``return` `(n * (n + 1) * (2 * n + 1)) / 6;``}` `// Driven Program``int` `main()``{``    ``int` `n = 4;``    ``cout << squaresum(n) << endl;``    ``return` `0;``}`

## 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.*/`

## Python3

 `# Python3 Program to``# find sum of square``# of first n natural``# numbers` `# Return the sum of``# square of first n``# natural numbers``def` `squaresum(n) :``    ``return` `(n ``*` `(n ``+` `1``) ``*` `(``2` `*` `n ``+` `1``)) ``/``/` `6` `# Driven Program``n ``=` `4``print``(squaresum(n))` `#This code is contributed by Nikita Tiwari.                                                              `

## C#

 `// C# Program to find sum``// of square of first n``// natural numbers``using` `System;` `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()` `    ``{``        ``int` `n = 4;``        ` `        ``Console.WriteLine(squaresum(n));``    ``}``}` `/*This code is contributed by vt_m.*/`

## PHP

 ``

## Javascript

 ``

Output :

`30`

Time Complexity: O(1)

Auxiliary Space: O(1), since no extra space has been taken

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.

## C++

 `// CPP Program to find sum of square of first``// n natural numbers. This program avoids``// overflow upto some extent for large value``// of n.``#include ``using` `namespace` `std;` `// Return the sum of square of first n natural``// numbers``int` `squaresum(``int` `n)``{``    ``return` `(n * (n + 1) / 2) * (2 * n + 1) / 3;``}` `// Driven Program``int` `main()``{``    ``int` `n = 4;``    ``cout << squaresum(n) << endl;``    ``return` `0;``}`

## Python3

 `# Python Program to find sum of square of first``# n natural numbers. This program avoids``# overflow upto some extent for large value``# of n.y` `def` `squaresum(n):``    ``return` `(n ``*` `(n ``+` `1``) ``/` `2``) ``*` `(``2` `*` `n ``+` `1``) ``/` `3` `# main()``n ``=` `4``print``(squaresum(n));` `# Code Contributed by Mohit Gupta_OMG <(0_o)>`

## 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)>`

## C#

 `// C# Program to find sum of square of first``// n natural numbers. This program avoids``// overflow upto some extent for large value``// of n.` `using` `System;` `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;``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 4;``        ` `        ``Console.WriteLine(squaresum(n));``    ``}``}` `// This Code is Contributed by vt_m.>`

## PHP

 ``

## Javascript

 ``

Output:

`30`

Time complexity: O(1) since performing constant operations

Space complexity: O(1) since using constant variables

My Personal Notes arrow_drop_up