Q. The coefficient of $x^{r}\lbrack 0 \leq r \leq (n - 1)\rbrack$ in the expansion of ${(x + 3)}^{n - 1} + {(x + 3)}^{n - 2}(x + 2) + (x + 3)^{n - 3}(x + 2)^{2} + ... + (x + 2)^{n - 1}$ is
A) $\ ^{n}C_{r}(3^{r} - 2^{n})$
B) $\ ^{n}C_{r}(3^{n - r} - 2^{n - r})$
C) $\ ^{n}C_{r}(3^{r} + 2^{n - 1})$
D) None of these
Explanation
(b)
We have,
${(x + 3)}^{n - 1} + {(x + 3)}^{n - 2}(x + 2) + (x + 3)^{n - 3}(x + 2)^{2} + ... + (x + 2)^{n - 1}$
$= \frac{(x + 3)^{n} - (x + 2)^{n}}{(x + 3) - (x + 2)} = (x + 3)^{n} - (x + 2)^{n}$
$\left( \because\frac{x^{n} - a^{n}}{x - a} = x^{n - 1} + x^{n - 2}a^{1} + x^{n - 3}a^{2} + ... + a^{n - 1} \right)$
Therefore, the coefficient of $x^{r}$ in the given expression
$=$coefficient of $x^{r}$ in ${\lbrack(x + 3)}^{n} - (x + 2)^{n}\rbrack$
$= \ ^{n}C_{r}3^{n - r} - \ ^{n}C_{r}2^{n - r}$
$= \ ^{n}C_{r}(3^{n - r} - 2^{n - r})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.