First slide

Binomial theorem for positive integral Index

Question

If $C_{0}, C_{1}, C_{2}, \dots, C_{n}$ denote the binomial coefficients i11 the expansion of $(1 + x)^{n}$ then $a C_{0} + (a + b) C_{1} + (a + 2 b) C_{2} + \dots + (a + n b) C_{n} =$

Moderate

A

$(a + n b) 2^{n}$
B

$(2 a + n b) 2^{n}$
C

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

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

Solution

We have,
$\begin{array}{l} a C_{0} + (a + b) C_{1} + (a + 2 b) C_{2} + \dots + (a + n b) C_{n} \\ = \sum_{r = 0}^{n} (a + r b)^{n} C_{r} \\ = \sum_{r = 0}^{n} a \cdot^{n} C_{r} + \sum_{r = 0}^{n} r b^{n} C_{r} \\ = a (\sum_{r = 0}^{n}^{n} C_{r}) + b (\sum_{r = 0}^{n} r \cdot^{n} C_{r}) \\ = a (\sum_{r = 0}^{n}^{n} C_{r}) + b (\sum_{r = 1}^{n} r \cdot \frac{n}{r} n - 1_{r}) \\ = a (\sum_{r = 0}^{n}^{n} C_{r}) + b n (\sum_{r = 1}^{n} n - 1) \end{array}$
$\begin{array}{l} = a \cdot 2^{n} + b n 2^{n - 1} [∵ \sum_{r = 0}^{n}^{n} C_{r} = 2^{n}, \sum_{r = 1}^{n}^{n - 1} C_{r - 1} = 2^{n - 1}] \\ = (2 a + b n) 2^{n - 1} \end{array}$

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