First slide

Binomial theorem for positive integral Index

Question

$The value of \frac{1}{(81)^{n}} - {}^{2 n}C_{1} \frac{10}{(81)^{n}} + {}^{2 n}C_{2} \frac{(10)^{2}}{(81)^{n}} \dots + \frac{(10)^{2 n}}{(81)^{n}} is$

Easy

A

${(-1)}^{n}$
B

0
C

1
D

none of these

Solution

$\begin{array}{l} G \cdot E = \frac{1}{(81)^{n}} [1 - {}^{2 n}C_{1} \cdot 10 + {}^{2 n}C_{2} (10)^{2} - \dots + {}^{2 n}C_{2 n} (10)^{2 n}] \\ = \frac{1}{(81)^{n}} (1 - 10)^{2 n} \\ = \frac{1}{(81)^{n}} (- 9)^{2 n} = 1 \end{array}$

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