Python Program for Sum of squares of first n natural numbers

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