Question

below o A double coer utePI(int eurber) 1) return (number comput PI(number number) return (number return (number return 1.0 1) return computePI er 1)s (number number) number 1) 1) number) equal the

Answer 1

The answer is B. Given a number 'n' we need to calculate the sum of the following series

$\frac{1}{1^{2}} + \frac{1}{2^{2}} +\frac{1}{3^{2}} +... +\frac{1}{n^{2}}$
which can further be simplified into the following form
$\frac{1}{1*1} + \frac{1}{2*2} +\frac{1}{3*3} +... +\frac{1}{n*n}$

To get the sum of the above series, we need to first find compute the value of $\frac{1}{n*n}$ and add it to the sum of the $\frac{1}{1^{2}} + \frac{1}{2^{2}} +\frac{1}{3^{2}} +... +\frac{1}{(n-1) * (n-1)}$ , which is upto (n-1)th term.

Now, to get the sum of the above series, we need to first compute the value of $\frac{1}{(n-1) * (n-1)}$ and it to the sum of the remaining series $\frac{1}{1^{2}} + \frac{1}{2^{2}} +\frac{1}{3^{2}} +... +\frac{1}{(n-2) * (n-2)}$ , which is upto (n-2)th term. And so on, we compute the value of the entire series, in a recursive manner.

The following function does the same thing as discussed above. It first computes the value of 1/(n * n) and add it to the sum of the series upto (n-1)th term, computed in a recursive manner. Each time you go on decrementing n by 1 until you reach 1. At this point you will return 1 and all the terms starts up getting added.

double computePI(int number) {

if (number <= 1) {
          return 1.0; // This is the base case
}
return 1 / (number * number ) + computePI(number - 1);

}