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
A.
3
B.
4
C.
2
D.
None of these
Answer :
2
Solution :
$$\eqalign{
& 1 + {99^n} = 1 + {\left( {100 - 1} \right)^n} = 1 + \left\{ {^n{C_0}{{100}^n} - {\,^n}{C_1} \cdot {{100}^{n - 1}} + ..... - {\,^n}{C_n}} \right\}\,\,{\text{because }}n{\text{ is odd}} \cr
& 1 + {99^n} = 100\left\{ {^n{C_0} \cdot {{100}^{n - 1}} - {\,^n}{C_1} \cdot {{100}^{n - 2}} + ..... - {\,^n}{C_{n - 2}} \cdot 100 + {\,^n}{C_{n - 1}}} \right\} \cr} $$
$$1 + {99^n} = 100 \times $$ integer whose units place is different from 0 ( $$n$$ having odd digit in units place).
Releted MCQ Question on Algebra >> Binomial Theorem
Releted Question 1
Given positive integers $$r > 1, n > 2$$ and that the co - efficient of $${\left( {3r} \right)^{th}}\,{\text{and }}{\left( {r + 2} \right)^{th}}$$ terms in the binomial expansion of $${\left( {1 + x} \right)^{2n}}$$ are equal. Then
If in the expansion of $${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n},$$ the co-efficients of $$x$$ and $${x^2}$$ are $$3$$ and $$- 6\,$$ respectively, then $$m$$ is