Question

If the coefficients of $x^{-7}$ and $x^{-8}$ to the expansion of${(2+\frac{1}{3x})}^n$ are equal then $n=$

• A. 56
• B. 15
• C. 45
• D. 55

Answer 1

The correct option is D 55

$(2+\frac{1}{3x}{)}^n$

$T_{r+1}{=}^n{C}_r.2^{n-r}.(3x{)}^{-r}$

such r = 7 , we get coefficients of $x^{-7}$

which is ${}^n{C}_7.\frac{2^{n-7}}{3^7}$

such r = 8 , we get coefficient of $x^{-8},$

which is ${}^n{C}_8.\frac{2^{n-8}}{3^8}$

given that both are equal.

${}^n{C}_7\frac{2^n}{3^7{2}^7}{=}^n{C}_8\frac{2^n}{2^8{.3}^8}$

${}^n{C}_7=\frac{{}^n{C}_8}{6}$

$\frac{n!}{7!(n-7)!}=\frac{n!}{6.8!(n-8)!}$

$\frac{8.7!(n-8)!}{7!(n-7)(n-8)!}=\frac{1}{6}$

$\frac{8}{n-7}=\frac{1}{6}$

n - 7 = 48

n = 55 .