First slide

Binomial theorem for positive integral Index

Question

If $(1 + x)^{15} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{15} x^{15}$ then value of the expression $S = C_{2} + 2 C_{3} + 3 C_{4} + \dots + 14 C_{15}$ is

Moderate

A

$13 (2^{14}) + 1$
B

$13 (2^{14})$
C

$13 (2^{15})$
D

none of these

Solution

We add $- 1 C_{0} + 0 C_{1}$ to both the sides of $S$ and let
$S_{1} = S - 1 C_{0} + 0 C_{1} = S - 1 .$
Note that
$S_{1} = - 1 C_{0} + 0 C_{1} + 1 C_{2} + 2 C_{3} + \dots + 13 C_{14} + 14 C_{15} (1)$
Using $C_{r} = C_{n - r}$ , we rewrite the above expression as
$S_{1} = 14 C_{0} + 13 C_{1} + 12 C_{2} + \dots + 0 C_{14} + (- 1) C_{15} (2)$
Adding $(1)$ and $(2)$ , we get
$\begin{array}{l} 2 S_{1} = 13 [C_{0} + C_{1} + C_{2} + \dots + C_{15}] = 13 (2^{15}) \\ \Rightarrow S_{1} = 13 (2^{14}) \Rightarrow S = 13 (2^{14}) + 1 \end{array}$

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