Question

If sum of the first 3 coefficients is $16$ in the expansion ${(x+\frac{1}{x^3})}^n$, then find n.

• A. $10$
• B. $8$
• C. $5$
• D. $4$

Answer 1

The correct option is C $5$

Given ${(x+\frac{1}{x^3})}^n$

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

$T_{r+1}={}^n{C}_r(x{)}^{n-r}{\left(\frac{1}{x^3}\right)}^r$

$T_{r+1}={}^n{C}_r(x{)}^{n-r-3r}$

$T_{r+1}={}^n{C}_r(x{)}^{n-4r}$

Given, sum of the first three coefficients is $16$

$T_1=T_{0+1}={}^n{C}_0(x{)}^n$

$T_2=T_{1+1}={}^n{C}_1(x{)}^{n-4}$

$T_3=T_{2+1}={}^n{C}_2(x{)}^{n-8}$

${}^n{C}_0+{}^n{C}_1+{}^n{C}_2=16$

$\frac{n!}{n!}+\frac{n!}{1!(n-1)!}+\frac{n!}{2!(n-2)!}=16$

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

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

$\frac{2n+n^2-n}{2}=15$

$n^2+n=30$

$n^2+n-30=0$

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

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

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

$\therefore$ Since n cannot be negative

so, $n=5$

Hence, option C is correct.