First slide

Binomial theorem for positive integral Index

Question

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

Moderate

A

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

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

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

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

Solution

$\begin{array}{l} (^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + \dots + (^{21} C_{10} -^{10} C_{10}) \\ =^{21} C_{1} +^{21} C_{2} + \dots +^{21} C_{10}) - (^{10} C_{1} +^{10} C_{2} + \dots +^{10} C_{10}) \\ = \frac{1}{2} (^{21} C_{1} +^{21} C_{2} + \dots +^{21} C_{20}) - (2^{10} - 1) \\ = \frac{1}{2} (^{21} C_{1} +^{21} C_{2} + \dots +^{21} C_{21} - 1) - (2^{10} - 1) \\ = \frac{1}{2} (2^{21} - 2) - (2^{10} - 1) = 2^{20} - 1 - 2^{10} + 1 \\ = 2^{20} - 2^{10} \end{array}$

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