Let `n` be an odd natural number greater than 1. Then , find the number of zeros at the end of the sum `99^n+1.`

Let n be an odd natural number of greater than 1. then the number of zeros at the end of the sum 999^(n)+1 is:

The number of zeros at the end of 99^(100) - 1 is -

Number of zeroes at the end of 99^(1001)+1?

The sum of n odd natural numbers is

If n is an odd number greater than 1, then n(n^(2)-1) is

Sum of 1 to n natural numbers is 36, then find the value of n.

If the sum of first n odd natural number is 1225, find the value of n.

If n is an odd natural number greater than 1 then the product of its successor and predecessor is an odd natural number (b) is an even natural number can be even or odd (d) None of these

The sum of first n odd natural numbers is