Program for Sum of squares of first n natural numbers

Given a positive integer n, we have to find the sum of squares of first n natural numbers. 
Examples :

Input : n = 2
Output: 5
Explanation: 1^2+2^2 = 5

Input : n = 8
Output: 204
Explanation : 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 = 204

[Naive Approach] - Adding One By One - O(n) Time and O(1) Space

The idea for this naive approach is to run a loop from 1 to n and sum up all the squares. 

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[Expected Approach]- Using Mathematical Formulae - O(1) Time and O(1) Space

The idea for this approach is to use the mathematical formulae for the sum of squares of first n natural numbers.

1² + 2² + ......... + n² = n(n+1)(2n+1) / 6

We can prove this formula using induction. We can easily see that the formula is true for n = 1 and n = 2 as sums are 1 and 5 respectively.

Let it be true for n = k-1. So sum of k-1 numbers
is (k – 1) * k * (2 * k – 1)) / 6

In the following steps, we show that it is true
for k assuming that it is true for k-1.

Sum of k numbers = Sum of k-1 numbers + k²
= (k – 1) * k * (2 * k – 1) / 6 + k²
= ((k² – k) * (2*k – 1) + 6k²)/6
= (2k³ – 2k² – k² + k + 6k²)/6
= (2k³ + 3k² + k)/6
= k * (k + 1) * (2*k + 1) / 6

Example : Find sum of squares of the first 3 natural numbers
Solution:
= 3 * (3 + 1) * (2*3 + 1) / 6
= (3 * 4 * 7) / 6
= 84 / 6
= 14

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Avoiding the overflow: 
In the above method, sometimes due to large value of n, the value of (n * (n + 1) * (2 * n + 1)) would overflow. We can avoid this overflow up to some extent using the fact that n*(n+1) must be divisible by 2 and restructuring the formula as (n * (n + 1) / 2) * (2 * n + 1) / 3;

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