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