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

The correct option is A Both (A) & (R) are individually true & (R) is correct explanation of (A).

$P\left(n\right)={99}^n+1=1+(100-1{)}^n=1{+}^n{C}_0{100}^n{-}^n{C}_1{100}^{n-1}+.....{+}^n{C}_{n-1}{100}^1{-}^n{C}_n(\because n=odd)$
$P\left(n\right)={}^n{C}_0\left(100{)}^n{-}^n{C}_1\right(100{)}^{n-1}+.....+{100}^n{C}_{n-1}$ $={10}^2{[}^n{C}_0{100}^{n-2}{-}^n{C}_1{100}^{n-3}+....{.}^n{C}_{n-1}]$

$=100\times$ integer whose unit place is different from zero (n having odd digit in the unit place).