Question

If the coefficients of $x^7$ and $x^8$ in $\left( 2 + \frac{x}{3} \right)^n$ are equal, then n is

• A. 56
• B. 55
• C. 49
• D. 50

Answer 1

Answer: (B)

55

Answer 2

Since $T_{r+1} =^{n}C_{r} a^{n-r}x^{r}$ in expansion of $\left(a+x\right)^{n}$ $T_{8} = {^{n}C_{7}} \left(2\right)^{n-7} \left(\frac{x}{3}\right)^{7} = {^{n}C_{7}} \frac{2^{n-7}}{3^{7}} x^{7}$ and $T_{9} = {^{n}C_{8}} \left(2\right)^{n-8} \left(\frac{x}{3}\right)^{8} = {^{n}C_{8}} \frac{2^{n-8}}{3^{8}} x^{8}$ Therefore, ${^{n}C_{7}} \frac{2^{n-7}}{3^{7} } = {^{n}C_{8}} \frac{2^{n-8}}{3^{8}}$ (since it is given that coefficient of $x^7$ = coefficient $x^8$) $\Rightarrow \frac{n!}{7! \left(n-7\right)!} \times \frac{8! \left(n-8\right)!}{n!} = \frac{2^{n-8}}{3^{8}} . \frac{3^{7}}{2^{n-7}}$ $\Rightarrow \frac{8}{n-7} = \frac{1}{6} \Rightarrow n = 55$