First slide

Binomial theorem for positive integral Index

Question

The sum of the series $^{20} C_{0} -^{20} C_{1} +^{20} C_{2} -^{20} C_{3} + \dots +^{20} C_{10}$ is

Moderate

A

$^{- 20} C_{10}$
B

${\frac{1}{2}}^{20} C_{10}$
C

0
D

$^{20} C_{10}$

Solution

We know that,
$(1 + x)^{20} =^{20} C_{0} +^{20} C_{1} x + \dots +^{20} C_{10} x^{1} + \dots +^{20} C_{20} x^{20}$ On putting x = -1, we get
$0 =^{20} C_{0} -^{20} C_{1} + \dots -^{20} C_{9} +^{20} C_{10} -^{20} C_{11} + \dots +^{20} C_{20}$
$\Rightarrow 0 =^{20} C_{0} -^{20} C_{1} + \dots -^{20} C_{9} +^{20} C_{10} -^{20} C_{9} + \dots +^{20} C_{0}$
$[∵^{n} C_{r} =^{n} C_{n - r}]$
$\Rightarrow 0 = 2 (^{20} C_{0} -^{20} C_{1} + \dots -^{20} C_{9}) +^{20} C_{10}$
$\begin{array}{l} \Rightarrow^{20} C_{10} = 2 (^{20} C_{0} -^{20} C_{1} + \dots +^{20} C_{10}) \\ \Rightarrow^{20} C_{0} -^{20} C_{1} + \dots +^{20} C_{10} \\ = {\frac{1}{2}}^{20} C_{10} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App