Test Index

Binomial Theorem Part 1

Section: Mathematics

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Question : 92 of 100

Marks: +1, -0

The coefficient of

x^n

in expansion of

(1+x)(1-x)^n

$(-1)^{n-1} n$
$(-1)^n(1-n)$
$(-1)^{n-1}(n-1)^2$
$(n-1)$

Solution:

Given

(1+x)(1-x)^n

=(1-x)^n+x(1-x)^n

General term of

(1-x)^n={}^n C_r \cdot (-1)^r \cdot x^r

\therefore

Term containing

x^n

in

(1-x)^n={}^n C_n \cdot (-1)^n \cdot x^n

So coefficient of

x^n

in

(1-x)^n={}^n C_n \cdot (-1)^n

General term of

x(1-x)^n={}^n C_r \cdot (-1)^r \cdot x^{r+1}

\therefore

Term containing

x^n

in

x(1-x)^n={}^n C_{n-1} \cdot (-1)^{n-1} \cdot x^n

So coefficient of

x^n

in

x(1-x)^n={}^n C_{n-1} \cdot (-1)^{n-1}

\therefore

coefficient of

x^n

in

(1-x)^n+x(1-x)^n

={}^n C_n \cdot (-1)^n+{}^n C_{n-1} \cdot (-1)^{n-1}

=(-1)^{n-1} [{}^n C_{n-1}-{}^n C_n]

=(-1)^{n-1}[n-1]

=(-1)^n[1-n]

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