Question

The numerically greatest term in the expansion of \((3-5x)^{11}\) when \(x=\dfrac{1}{5}\), is

• A. \(55\times 3^9\)
• B. \(55\times 3^6\)
• C. \(45\times 3^9\)
• D. \(45\times 3^6\)

Hint:

For numerically greatest term, use ratio \(\dfrac{T_{r+2}}{T_{r+1}}\) and find the \(r\) where it changes from \(>1\) to \(<1\).

Answer 1

The Correct Option is A

Step 1: Write general term.
In expansion of \((3-5x)^{11}\), general term is:
\[ T_{r+1} = \binom{11}{r}(3)^{11-r}(-5x)^r \] Step 2: Substitute \(x=\dfrac{1}{5}\).
\[ -5x = -5\left(\frac{1}{5}\right) = -1 \] So term becomes:
\[ T_{r+1} = \binom{11}{r}3^{11-r}(-1)^r \] Numerical value is:
\[ |T_{r+1}| = \binom{11}{r}3^{11-r} \] Step 3: Find greatest term using ratio.
\[ \frac{|T_{r+2}|}{|T_{r+1}|} = \frac{\binom{11}{r+1}3^{10-r}}{\binom{11}{r}3^{11-r}} = \frac{11-r}{r+1}\cdot \frac{1}{3} \] We need greatest term where ratio just becomes less than 1.
\[ \frac{11-r}{r+1}\cdot \frac{1}{3} \le 1 \Rightarrow 11-r \le 3(r+1) \Rightarrow 11-r \le 3r+3 \Rightarrow 8 \le 4r \Rightarrow r \ge 2 \] Check for \(r=1\):
\[ \frac{11-1}{2}\cdot\frac{1}{3} = \frac{10}{6}>1 \] So terms increasing till \(r=2\).
Thus greatest term is at \(r=2\), i.e. \(T_{3}\).
Step 4: Compute \(T_3\).
\[ T_3 = \binom{11}{2}3^{9}(-1)^2 = 55\cdot 3^9 \] Final Answer: \[ \boxed{55\times 3^9} \]