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

Last Updated : 25 Apr, 2023

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

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

## 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.`

Output

`30`

Time complexity: O(1)
Auxiliary space: O(1)

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.

## 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)>`

Output

`30.0`

Time complexity: O(1) because constant operations are being performed
Auxiliary space: O(1)

## Approach: list comprehension

1. Take input from the user as a positive integer N using the input() function.
2. Convert the input string to an integer using the int() function and store it in a variable N.
3. Use a list comprehension to create a list of squares of numbers from 1 to N. The list comprehension should look like this: [i*i for i in range(1, N+1)]. This creates a list of squares of numbers from 1 to N.
4. Use the sum() function to find the sum of all elements in the list. Store the result in a variable sum_of_squares.
5. Print the result using the print() function.

## Python3

 `# Take input from user` `N ``=` `5`   `# Use list comprehension to create list of squares` `squares_list ``=` `[i``*``i ``for` `i ``in` `range``(``1``, N``+``1``)]`   `# Find sum of squares using sum() function` `sum_of_squares ``=` `sum``(squares_list)`   `# Print the result` `print``(``"Sum of squares of first"``, N, ``"natural numbers is"``, sum_of_squares)`

Output

`Sum of squares of first 5 natural numbers is 55`

The time complexity using a list comprehension is O(n)

The auxiliary space or space complexity is also O(n)

## Approach: iterative

We can use a loop to iterate through the first n natural numbers and calculate the sum of their squares.

Steps:

1. Initialize a variable sum to 0.
2. Use a loop to iterate through the first n natural numbers, i.e., from 1 to n.
3. Within the loop, calculate the square of the current number and add it to the sum.
4. After the loop completes, print the value of sum.

## Python3

 `n ``=` `4` `sum` `=` `0` `for` `i ``in` `range``(``1``, n``+``1``):` `    ``sum` `+``=` `i``*``*``2`   `print``(``"The sum of squares of first"``, n, ``"natural numbers is"``, ``sum``)`

Output

`The sum of squares of first 4 natural numbers is 30`

Time Complexity: O(n)
Auxiliary Space: O(1)

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

METHOD 5:Using recursion

APPROACH:

This program calculates the sum of squares of first n natural numbers recursively.

ALGORITHM:

1.Define a function sum_of_squares(n) which takes an integer n as input.
2.Check if n is 1, return 1.
3.Else, calculate the sum of squares recursively by adding n*n with the sum_of_squares of n-1.
4.Set the value of N as 4.
5.Call sum_of_squares function with N as input and store the result in sum_of_squares variable.
6.Print the result.

## Python3

 `def` `sum_of_squares(n):` `    ``if` `n ``=``=` `1``:` `        ``return` `1` `    ``else``:` `        ``return` `n``*``n ``+` `sum_of_squares(n``-``1``)`   `N ``=` `4` `sum_of_squares ``=` `sum_of_squares(N)`   `print``(``"Sum of squares of first"``, N, ``"natural numbers:"``, sum_of_squares)`

Output

`Sum of squares of first 4 natural numbers: 30`

Time Complexity: O(n)
Space Complexity: O(n)