SATHEE: Binomial Theorem 1 Question 15

Binomial Theorem 1 Question 15

16. 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

(a) $2^{21} - 2^{11}$

(b) $2^{21} - 2^{10}$

(d) $2^{20} - 2^{10}$

(2017 Main)

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Answer:

Correct Answer: 16. $(101)^{50}$ 34. $\frac{2^{m n} - 1}{2^{m n} (2^{n} - 1)}$

Solution:

$(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3})$

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

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Binomial Theorem 1 Question 15

16. The value of (21C1−10C1)+(21C2−10C2) +(21C3−10C3)+(21C4−10C4)+…+(21C10−10C10) is

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16. 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