First slide

Binomial theorem for positive integral Index

Question

The value of $-^{15} C_{1} + 2 \cdot^{15} C_{2} - 3 \cdot^{15} C_{3} + \dots . 15 \cdot^{15} C_{15}$ $+^{14} C_{1} +^{14} C_{3} +^{14} C_{5} + \dots +^{14} C_{11}$ is

Moderate

A

2¹⁴
B

2¹³-13
C

2^16-1
D

2¹³-14

Solution

$\begin{array}{l} S_{1} =^{15} C_{1} + 2 \cdot^{15} C_{2} - \dots \dots - 15^{15} C_{15} \\ = \sum_{r = 1}^{15} (- 1)^{r} \cdot r^{15} C_{2} = 15 \sum_{r = 1}^{15} {(- 1)^{r}}^{14} C_{r - 1} \\ = 15 (-^{14} C_{0} +^{14} C_{1} - \dots \dots -^{14} C_{14}) = 15 (0) = 0 \\ S_{2} =^{14} {C_{1}}^{14} C_{3} + \dots . . +^{14} C_{11} \\ = (^{14} C_{1} +^{14} C_{3} + \dots \dots .^{14} C_{11} +^{14} C_{13}) -^{14} C_{13} \\ = 2^{13} - 14 \\ = S_{1} + S_{2} = 2^{13} - 14 \end{array}$

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