Related Articles
Python Program for Sum of squares of first n natural numbers
• Difficulty Level : Medium
• Last Updated : 03 Dec, 2018

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.

 `# 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.*/`

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
```
 `# 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.                                                               `

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.

 `# 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)>`

Output:

```30
```

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

My Personal Notes arrow_drop_up