Binomial Theorem 1 Question 15
16.
The value of $\left({ }^{21} C_{1}-{ }^{10} C_{1}\right)+\left({ }^{21} C_{2}-{ }^{10} C_{2}\right)$ $+\left({ }^{21} C_{3}-{ }^{10} C_{3}\right)+\left({ }^{21} C_{4}-{ }^{10} C_{4}\right)+\ldots+\left({ }^{21} C_{10}-{ }^{10} C_{10}\right)$ is
(a) $2^{21}-2^{11}$
(b) $2^{21}-2^{10}$
(c) $2^{20}-2^{9}$
(d) $2^{20}-2^{10}$
(2017 Main)
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Answer:
Correct Answer: 16. (d)
Solution:
- $\left({ }^{21} C_{1}-{ }^{10} C_{1}\right)+\left({ }^{21} C_{2}-{ }^{10} C_{2}\right)+\left({ }^{21} C_{3}-{ }^{10} C_{3}\right)$
$ \begin{aligned} & \quad+\ldots+\left({ }^{21} C_{10}-{ }^{10} C_{10}\right) \\ & =\left({ }^{21} C_{1}+{ }^{21} C_{2}+\ldots+{ }^{21} C_{10}\right)-\left({ }^{10} C_{1}+{ }^{10} C_{2}+\ldots+{ }^{10} C_{10}\right) \\ & =\frac{1}{2}\left({ }^{21} C_{1}+{ }^{21} C_{2}+\ldots+{ }^{21} C_{20}\right)-\left(2^{10}-1\right) \\ & =\frac{1}{2}\left({ }^{21} C_{1}+{ }^{21} C_{2}+\ldots+{ }^{21} C_{21}-1\right)-\left(2^{10}-1\right) \\ & =\frac{1}{2}\left(2^{21}-2\right)-\left(2^{10}-1\right)=2^{20}-1-2^{10}+1=2^{20}-2^{10} \end{aligned} $
