Question

If sum of the coefficients of the first two odd terms of the expansion $(x+y{)}^n$ is 16 then find n.

• A. 10
• B. 8
• C. 7
• D. 6

Answer 1

The correct option is D 6

Given, $(x+y{)}^n$

General term can be written as

$T_{r+1}={}^n{C}_r{x}^{n-r}{y}^r$

given,

sum of coefficients of the first two odd terms is $16$

$T_1=T_{0+1}={}^n{C}_0{x}^{n-0}{y}^0$

$T_1={}^n{C}_0{x}^n$

$T_3=T_{2+1}={}^n{C}_2{x}^{n-2}{y}^2$

Sum of coeff of $T_1$ and $T_3=16$

${}^n{C}_0+{}^n{C}_2=16$

$\frac{n!}{n!}+\frac{n!}{2!(n-2)!}=16$

$\frac{n(n-1)(n-2)!}{2(n-2)!}=15$

$n(n-1)=30$

$n^2-n-30=0$

On solving $n^2-6n+5n-30=0$

$n(n-6)+5(n-6)=0$

$(n-6)(n+5)=0$

Since n is positive

$n=6$

Option D is correct.