SATHEE: Binomial Theorem 2 Question 10

Binomial Theorem 2 Question 10

13.

Prove that

$\begin{aligned} + (- 1)^{k} \begin{array}{cc} n & n - k \\ k & 0 \end{array} = \begin{matrix} n \\ k \end{matrix} (2003, 4 M) \end{aligned}$

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Solution:

To show that

$\begin{aligned} 2^{k} \cdot^{n} C_{0} \cdot^{n} C_{k} - 2^{k - 1} \cdot^{n} C_{1} \cdot^{n - 1} C_{k - 1} \\ + 2^{k - 2} \cdot^{n} C_{2} \cdot^{n - 2} C_{k - 2} - \dots + (- 1)^{k}^{n} C_{k}^{n - k} C_{0} =^{n} C_{k} \end{aligned}$

Taking LHS

$2^{k} \cdot^{n} C_{k}^{n} C_{k} - 2^{k - 1} \cdot^{n} C_{1} \cdot^{n - 1} C_{k - 1} + \dots + (- 1)^{k} \cdot^{n} C_{k} \cdot^{n - k} C_{0}$

$= \sum_{r = 0}^{k} (- 1)^{r} \cdot 2^{k - r} \cdot^{n} C_{r} \cdot^{n - r} C_{k - r}$

$= \sum_{r = 0}^{k} (- 1)^{r} 2^{k - r} \cdot \frac{n!}{r! (n - r)!} \cdot \frac{(n - r)!}{(k - r)! (n - k)!}$

$= \sum_{r = 0}^{k} (- 1)^{r} \cdot 2^{k - r} \cdot \frac{n!}{(n - k)! \cdot k!} \cdot \frac{k!}{r! (k - r)!}$

$= \sum_{r = 0}^{k} (- 1)^{r} \cdot 2^{k - r}^{n} C_{k} \cdot^{k} C_{r} = 2^{k} \cdot^{n} C_{k} [\sum_{r = 0}^{k} (- 1)^{r} \cdot \frac{1}{2^{r}} \cdot^{k} C_{r}]$

$= 2^{k} \cdot^{n} C_{k} (1 - \frac{1}{2})^{k} \overset{}{n} C_{k} = RHS$

Binomial Theorem 2 Question 10

13.

Solution:

Engineering Preparation

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Binomial Theorem 2 Question 10

13.

Solution:

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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