Question

Assertion (A): Let $n$ be an odd natural number greater than 1. Then the number of zero at the end of $P\left(n\right)=99^n+1$ is $2$.
Reason (R) : $99^n+1=10^2\times$ integer, whose unit place is different from zero.

• A. Both (A) & (R) are individually true & (R) is correct explanation of (A).
• B. Both (A) & (R) are individually true but (R) is not the correct (proper) explanation of (A).
• C. (A) is true but (R) is false.
• D. (A) is false but (R) is true

Answer 1

Answer: (A)

$P\left(n\right)=99^n+1,n$ is odd
$=1+{\left(100-1\right)}^n$
$=1{+}^n{C}_{0}100^n{-}^n{C}_{1}100^{n-1}+....{+}^n{C}_{n-1}{\left(100\right)}^1{-}^n{C}_n$ ($\because n=$ odd)
${=}^n{C}_0{\left(100\right)}^n{-}^n{C}_1{\left(100\right)}^{n-1}+...+100^n{C}_{n-1}$
$=10^2\left[{}^n{C}_{0}100^{n-2}{-}^n{C}_{1}100^{n-3}+...{+}^n{C}_{n-1}\right]$
$=100\times$ integer whose unit place is different from zero