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# 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

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