Question:
The coefficient of $x^{12}$ in the expansion of
\[
(3+2x)^{-5}
\]
is
The coefficient of $x^{12}$ in the expansion of
\[
(3+2x)^{-5}
\]
is
Show Hint
Always rewrite generalized binomial expressions into the form:
\[
(1+x)^n
\]
before applying the theorem.
Updated On: Jun 17, 2026
- ${}^{17}C_5\frac{3^{12}}{2^5}$
- ${}^{16}C_{12}\frac{2^{12}}{3^{17}}$
- ${}^{16}C_{12}\frac{2^{17}}{3^{12}}$
- ${}^{17}C_5\frac{3^{12}}{2^{17}}$
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
Using generalized binomial theorem:
\[
(1+x)^{-n}
=
1-nx+\frac{n(n+1)}{2!}x^2-\cdots
\]
General term:
\[
T_{r+1}
=
(-1)^r
{}^{n+r-1}C_r
x^r
\]
Step 1: Rewrite the expression.
\[ (3+2x)^{-5} = 3^{-5} \left( 1+\frac{2x}{3} \right)^{-5} \]
Step 2: Apply generalized binomial theorem.
General term: \[ T_{r+1} = 3^{-5} (-1)^r {}^{r+4}C_r \left( \frac{2x}{3} \right)^r \]
Step 3: Find coefficient of $x^{12}$.
For: \[ x^{12}, \quad r=12 \] Hence coefficient: \[ 3^{-5} {}^{16}C_{12} \left( \frac{2}{3} \right)^{12} \] \[ = {}^{16}C_{12} \frac{2^{12}}{3^{17}} \] Hence, \[ \boxed{ {}^{16}C_{12} \frac{2^{12}}{3^{17}} } \]
Step 1: Rewrite the expression.
\[ (3+2x)^{-5} = 3^{-5} \left( 1+\frac{2x}{3} \right)^{-5} \]
Step 2: Apply generalized binomial theorem.
General term: \[ T_{r+1} = 3^{-5} (-1)^r {}^{r+4}C_r \left( \frac{2x}{3} \right)^r \]
Step 3: Find coefficient of $x^{12}$.
For: \[ x^{12}, \quad r=12 \] Hence coefficient: \[ 3^{-5} {}^{16}C_{12} \left( \frac{2}{3} \right)^{12} \] \[ = {}^{16}C_{12} \frac{2^{12}}{3^{17}} \] Hence, \[ \boxed{ {}^{16}C_{12} \frac{2^{12}}{3^{17}} } \]
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