SATHEE: Binomial Theorem 2 Question 7

Binomial Theorem 2 Question 7

10.

If $C_{r}$ stands for $^{n} C_{r}$ , then the sum of the series

$\frac{2 (\frac{n}{2})! (\frac{n}{2})!}{n!} [C_{0}^{2} - 2 C_{1}^{2} + 3 C_{2}^{2} - \dots + (- 1)^{n} (n + 1) C_{n}^{2}]$

where $n$ is an even positive integer, is

(a) $(- 1)^{n / 2} (n + 2)$

(b) $(- 1)^{n} (n + 1)$

(d) None of these

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Answer:

Correct Answer: 10. (a)

Solution:

We have,

$\begin{aligned} C_{0}^{2} - 2 & C_{1}^{2} + 3 C_{2}^{2} - 4 C_{3}^{2} + \dots + (- 1)^{n} (n + 1) C_{n}^{2} \\ = & [C_{0}^{2} - C_{1}^{2} + C_{2}^{2} - C_{3}^{2} + \dots + (- 1)^{n} C_{n}^{2}] \\ - [C_{1}^{2} - 2 C_{2}^{2} + 3 C_{3}^{2} - \dots + (- 1)^{n} n C_{n}^{2}] \\ = & (- 1)^{n / 2} \frac{n!}{(\frac{n}{2})! (\frac{n}{2})!} - (- 1)^{\frac{n}{2} - 1} \frac{n}{2} \frac{n!}{(\frac{n}{2})! (\frac{n}{2})!} \\ = & (- 1)^{n / 2} \frac{n!}{(\frac{n}{2})! (\frac{n}{2})!} (1 + \frac{n}{2}) \end{aligned}$

$∴ \frac{2 (\frac{n}{2})! (\frac{n}{2})!}{n!} [C_{0}^{2} - 2 C_{1}^{2} + 3 C_{2}^{2} - \dots + (- 1)^{r} (n + 1) C_{n}^{2}]$

$= \frac{2 (\frac{n}{2})! (\frac{n}{2})!}{n!} (- 1)^{n / 2} \frac{n!}{(\frac{n}{2})! (\frac{n}{2})!} \frac{(n + 2)}{2} = (- 1)^{n / 2} (n + 2)$

Binomial Theorem 2 Question 7

10.

Answer:

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Binomial Theorem 2 Question 7

10.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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