The coefficient of x^ 5 in the expansion of (1 + x)21 + (1 + x)22 + + (1 + x)30 is

Correct answer: C. C6 31 - C6 21 (1 + x) ^ 21 + (1 + x) ^ 22 + + (1 + x) ^ 30 = (1 + x) ^ 21 [1 + (1 + x) ^ 1 + + (1 + x) ^ 9 ] = (1 + x) ^ 21 [ (1 + x) ^ 10 - 1 (1 + x) - 1 ] = (1)/(x)[ (1 + x) ^ 31 - (1 + x) ^ 21 ] ∴ Coefficient of x ^ 5 in the given expression = Coefficient of x ^ 5 in (1)/(x)[ (1 + x) ^ 31 - (1 + x) ^ 21 ] = Coefficient of x ^ 6 in [ (1 + x) ^ 31 - (1 + x) ^ 21 ] = C6 31 - C6 21

BITSAT 2017

Exam: BITSAT
Year: 2017
Subject: Mathematics
Chapter: Binomial Theorem

BITSAT 2017 · Mathematics · Binomial Theorem

The coefficient of x^ 5 in the expansion of (1 + x)21 + (1 + x)22 +...

The coefficient of

x^{5} in the expansion of

\left(1 + x\right)21 +

\left(1 + x\right)22 + \ldots + \left(1 + x\right)30

Options

A $C5 51$
B $C5 9$
C $C6 31 - C6 21$
D $C5 30 + C5 20$

Correct answer

C. $C6 31 - C6 21$

Step-by-step solution

${\left(1 + x\right)}^{21} + {\left(1 + x\right)}^{22} + \ldots \ldots + {\left(1 + x\right)}^{30}$

$= {\left(1 + x\right)}$ ^{21}\left[1 + {\left(1 + x\right)}^{1} + \ldots + {\left(1 + x\right)}^{9}\right] $$ = {\left(1 + x\right) $}^{21}\left[\dfrac{{\left(1 + x\right)}^{10} - 1}{\left(1 + x\right)$ - 1}\right] $= \dfrac{1}{x}\left[{\left(1 + x\right)}$ ^{31} - {\left(1 + x\right)}^{21}\right] $$ ∴ $Coefficient of$ {x}^{5} in the given expression

= Coefficient of ${x}^{5} in \dfrac{1}{x}\left[{\left(1 + x\right)}$ ^{31} - {\left(1 + x\right) $}^{21}\right]$

= Coefficient of ${x}^{6} in \left[{\left(1 + x\right)}^{31} - {\left(1 + x\right)}^{21}\right]$

$= C6 31 - C6 21$

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