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Q.
Let $n$ be an odd natural number greater than $1$ . Then, the number of zeros at the end of the sum $99^{n}+1$ is
Binomial Theorem
Solution:
$1+99 n=1+(100-1) n=1+\left\{n C_{0} 100 n-n C_{1} \cdot 100 n^{-1}+\ldots .\right.
n C n\}$ because $n$ is odd
$=100\left\{n C_{0} .100 n^{-1}-n C_{1} \cdot 100 n^{-2}+\ldots-n C n_{-2} .100+n C n_{-1}\right\}$
$=100 \times$ integer whose unit's place is different from $0$ .
$\left[\because n C n_{-1}=n\right.$, has odd digit at unit's place $]$
$\therefore $ There are two zeros at the end of the sum $99 n+1$.