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Binomial Theorem Part 1

Section: Mathematics

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Question : 25 of 100

Marks: +1, -0

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}

[24 Feb 2021 Shift 1]

$2^{16}-1$
$2^{13}-14$
$2^{13}-13$
$2^{14}$

Solution:

Given,

\left(-{}^{15}C_{1}+2\cdot{}^{15}C_{2}-3\cdot{}^{15}C_{3}+\dots-15\cdot{}^{15}C_{15}\right)

\;\;+\left({}^{14}C_{1}+{}^{14}C_{3}+{}^{14}C_{5}+\dots+{}^{14}C_{11}\right)

Let

S_{1}=-{}^{15}C_{1}+2\cdot{}^{15}C_{2}-3\cdot{}^{15}C_{3}+\dots-15\cdot{}^{15}C_{15}

\;\;=\sum\limits_{r=1}^{15}(-1)^{r}\cdot r\cdot{}^{15}C_{r}=15\sum\limits_{r=1}^{15}(-1)^{r}{}^{14}C_{r-1}

\;\;=15\left(-{}^{14}C_{0}+{}^{14}C_{1}-{}^{14}C_{2}+\dots-{}^{14}C_{14}\right)

\;\;=15(0)=0

S_{2}={}^{14}C_{1}+{}^{14}C_{3}+{}^{14}C_{5}+\dots+{}^{14}C_{11}

\;\;=\left({}^{14}C_{1}+{}^{14}C_{3}+\dots+{}^{14}C_{13}\right)-{}^{14}C_{13}

\;\;=2^{13}-14

Now, the required value is

\left(-{}^{15}C_{1}\;+2\cdot{}^{15}C_{2}-3\cdot{}^{15}C_{3}+\dots-15\cdot{}^{15}C_{15}\right)

\;+\left({}^{14}C_{1}+{}^{14}C_{3}+\dots+{}^{14}C_{11}\right)

\;\;=S_{1}+S_{2}=0+2^{13}-14=2^{13}-14

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