If the coefficients of \(x^9\) and \(x^{10}\) in the binomial expansion of \(\left(3+\frac{x}{2}\right)^n\)…

If the coefficients of \(x^9\) and \(x^{10}\) in the binomial expansion of \(\left(3+\frac{x}{2}\right)^n\) are equal, then \(n=\)
  1. 69
  2. 96
  3. 66
  4. 99

Solution

Given binomial is \(\left(3+\frac{x}{2}\right)^n\) and the general term in the expansion is \({ }^n C_r \frac{3^{n-r}}{2^r} x^r\) \(\therefore\) Coefficient of \(x^9\) and \(x^{10}\) are \({ }^n C_9 \frac{3^{n-9}}{2^9} \text { and }{ }^n C_{10} \frac{3^{n-10}}{2^{10}} \text { respectively }\) \(\because\) It is given that, the coefficients \(x^9\) and \(x^{10}\) are equal, \(\begin{aligned} & \text {So, } { }^n C_9 \frac{3^{n-9}}{2^9} ={ }^n C_{10} \frac{3^{n-10}}{2^{10}} \\ & \Rightarrow \frac{n ! \times 3}{9 !(n-9) !} =\frac{n !}{10 !(n-10) ! \times 2} \\ \Rightarrow & \frac{3}{n-9} =\frac{1}{10 \times 2} \Rightarrow n-9=60 \\ \Rightarrow & n =69 \end{aligned}\) Hence, option (a) is correct.

Asked in: AP EAMCET 2020 (21 Sep Shift 1)

Practice more Binomial Theorem questions on Aicharya