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

1
C

0
D

none

Solution

We have $\frac{1}{{(81)}^{n}} [1 -^{2 n} C_{1} .10 +^{2n} C_{2} {(10)}^{2} - ... +^{2 n} C_{2 n} {(10)}^{2 n}] = \frac{1}{{(81)}^{n}} {(1 - 10)}^{2n} = \frac{1}{{(81)}^{n}} {(- 9)}^{2n} = 1$

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