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 = 4print(squaresum(n)) # |
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 = 4print(squaresum(n)) # |
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 = 4print(squaresum(n)); # |
Output:
30