Guest

nC0 2nCn - nC1 (2n-2)Cn + nC2 (2n-4)Cn.............. equals to

nC0 2nCn - nC1 (2n-2)Cn + nC2 (2n-4)Cn.............. equals to 

Grade:11

1 Answers

Ayush Raj
11 Points
6 years ago
$({^nC_0})({^{2n}C_n})-({^nC_1})({^{2n-2}C_n})+({^nC_2})({^{2n-4}C_n}).....$
$=$the coefficient of $x^n$ in $[{^nC_0(1+x)^{2n}} -{^nC_1(1+x)^{2n-2}}+{^nC_2(1+x)^{2n-4}}.....]$
$=$coefficient of $x^n$ in $[{^nC_0((1+x)^2)^{n}} -{^nC_1((1+x)^2)^{n-1}}+{^nC_2((1+x)^2)^{n-2}}.....]$
$=$coefficient of $x^n$ in $[(1+x)^2-1]^n$
$=$coefficient of $x^n$ in $(2x+x^2)^n$
$=$coefficient of $x^n$ in $x^n(2+x)^n$
$=$ $2^n$
Note that in such questions,you’ll have to take hint from the question-so to choose the right power of x whose coefficient gives the answer.

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free