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.

Write the sequence of odd numbers greater than 1.

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