Question

In the binomial expansion of \((1+x)^n\), the coefficients of \(x^{k-1}\), \(x^k\), \(x^{k+1}\) and also the coefficients of \(x^{l-1}\), \(x^l\), \(x^{l+1}\) are in A.P.

Then, the numerically greatest term in the expansion of \((1+x)^{14}\) when \(x=\dfrac{2}{3}\) is

• A. \({}^{14}C_{5}\left(\dfrac23\right)^5\)
• B. \({}^{14}C_{4}\left(\dfrac23\right)^4\)
• C. \({}^{15}C_{7}\left(\dfrac23\right)^8\)
• D. \({}^{15}C_{6}\left(\dfrac23\right)^9\)

Hint:

For the expansion of \((1+x)^n\), the greatest term is obtained by comparing successive terms using \[ \frac{T_{r+2}}{T_{r+1}} =\frac{n-r}{r+1}\,x. \] The last term for which this ratio is at least \(1\) is the numerically greatest term.

Answer 1

The Correct Option is A

Step 1: Use the greatest term criterion. The \((r+1)^{\text{th}}\) term of \[ (1+x)^{14} \] is \[ T_{r+1} =\binom{14}{r}\left(\frac23\right)^r. \] The ratio of consecutive terms is \[ \frac{T_{r+2}}{T_{r+1}} =\frac{14-r}{r+1}\cdot\frac23. \]

Step 2: Find the largest term. For the greatest term, \[ \frac{T_{r+2}}{T_{r+1}}\ge1. \] Thus, \[ \frac{14-r}{r+1}\cdot\frac23\ge1 \] \[ 2(14-r)\ge3(r+1) \] \[ 28-2r\ge3r+3 \] \[ 25\ge5r \] \[ r\le5. \] Hence, \[ T_6 \] is the numerically greatest term.

Step 3: Write the greatest term. Therefore, \[ T_6 = \binom{14}{5} \left(\frac23\right)^5. \] Hence, \[ \boxed{{}^{14}C_{5}\left(\frac23\right)^5}. \] Thus, \[ \boxed{(A)} \] is the correct answer.