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. $(101)^{50}$ 34. $\frac{2^{m n}-1}{2^{m n}\left(2^{n}-1\right)}$
Solution:
- $\left({ }^{21} C _1-{ }^{10} C _1\right)+\left({ }^{21} C _2-{ }^{10} C _2\right)+\left({ }^{21} C _3-{ }^{10} C _3\right)$
$$ \begin{aligned} & \quad+\ldots+\left({ }^{21} C _{10}-{ }^{10} C _{10}\right) \\ & =\left({ }^{21} C _1+{ }^{21} C _2+\ldots+{ }^{21} C _{10}\right)-\left({ }^{10} C _1+{ }^{10} C _2+\ldots+{ }^{10} C _{10}\right) \\ & =\frac{1}{2}\left({ }^{21} C _1+{ }^{21} C _2+\ldots+{ }^{21} C _{20}\right)-\left(2^{10}-1\right) \\ & =\frac{1}{2}\left({ }^{21} C _1+{ }^{21} C _2+\ldots+{ }^{21} C _{21}-1\right)-\left(2^{10}-1\right) \\ & =\frac{1}{2}\left(2^{21}-2\right)-\left(2^{10}-1\right)=2^{20}-1-2^{10}+1=2^{20}-2^{10} \end{aligned} $$
