First slide

Binomial theorem for positive integral Index

Question

The value of $(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2})$ $+ (^{21} C_{3} -^{10} C_{3}) + (^{21} C_{4} -^{10} C_{4}) + \dots + + (^{21} C_{10} -^{10} C_{10})$ is

Moderate

A

$2^{21} - 2^{11}$
B

$2^{21} - 2^{10}$
C

$2^{20} - 2^{9}$
D

$2^{20} - 2^{10}$

Solution

We have,
$\begin{array}{l} (^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) \\ + (^{21} C_{4} -^{10} C_{4}) + \dots + (^{21} C_{10} -^{10} C_{10}) \end{array}$
$\begin{array}{l} (^{21} C_{1} +^{21} C_{2} +^{21} C_{3} +^{21} C_{4} + \dots +^{21} C_{10}) \\ - (^{10} C_{1} +^{10} C_{2} +^{10} C_{3} +^{10} C_{4} + \dots +^{10} C_{10}) \\ = \frac{1}{2} \{2^{21} C_{1} + 2^{21} C_{2} + 2^{21} C_{3} + 2^{21} C_{4} + \dots + 2^{21} C_{10}\} \\ - \{^{10} C_{1} +^{10} C_{2} +^{10} C_{3} +^{10} C_{4} + \dots +^{10} C_{10}\} \end{array}$
$\begin{array}{l} = (^{21} C_{1} +^{21} C_{2} +^{21} C_{3} +^{21} C_{4} + \dots +^{21} C_{10}) \\ - (^{10} C_{1} +^{10} C_{2} +^{10} C_{3} +^{10} C_{4} + \dots +^{10} C_{10}) \\ = \frac{1}{2} \{2^{21} C_{1} + 2^{21} C_{2} + 2^{21} C_{3} + 2^{21} C_{4} + \dots + 2^{21} C_{10}\} \end{array}$
$\begin{array}{l} = \frac{1}{2} \{(^{21} C_{1} +^{21} C_{20}) + (^{21} C_{2} +^{21} C_{19}) + (^{21} C_{3} +^{21} C_{18}) \\ + \dots + (^{21} C_{10} +^{21} C_{11})\} - \{^{10} C_{1} +^{10} C_{2} \end{array}$
$\begin{array}{l} = \frac{1}{2} \{2^{21} - (^{21} C_{0} +^{21} C_{21})\} - \{2^{10} -^{10} C_{0}\} \\ = \frac{1}{2} (2^{21} - 2) - (2^{10} - 1) = 2^{20} - 2^{10} \end{array}$

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