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. 

Below is the implementation of this approach  

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 * ? k2 + 3 * ? k + ? 1
n3 + 3 * n2 + 3 * n + 1 = 1 + 3 * ? k2 + 3 * (n * (n + 1))/2 + n
n3 + 3 * n2 + 3 * n = 3 * ? k2 + 3 * (n * (n + 1))/2 + n
n3 + 3 * n2 + 2 * n - 3 * (n * (n + 1))/2 = 3 * ? k2
n * (n2 + 3 * n + 2) - 3 * (n * (n + 1))/2 = 3 * ? k2
n * (n + 1) * (n + 2) - 3 * (n * (n + 1))/2 = 3 * ? k2
n * (n + 1) * (n + 2 - 3/2) = 3 * ? k2
n * (n + 1) * (2 * n + 1)/2 = 3 * ? k2
n * (n + 1) * (2 * n + 1)/6 = ? k2

Below is the implementation of this approach: 

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. 

Output:

30

Time complexity: O(1) since performing constant operations

Space complexity: O(1) since using constant variables