Sequence And Series Question 569
Question: The coefficient of $ x^{r} $ in the expansion of $ 1+\frac{a+bx}{1,!}+\frac{{{(a+bx)}^{2}}}{2,!}+…..+\frac{{{(a+bx)}^{n}}}{n,!}+….. $ is
[MP PET 1989]
Options:
A) $ \frac{{{(a+b)}^{r}}}{r,!} $
B) $ \frac{b^{r}}{r,!} $
C) $ \frac{e^{a}b^{r}}{r,!} $
D) $ {e^{a+b^{r}}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ S={e^{a+bx}}=e^{a}.e^{bx}=e^{a}{ 1+\frac{bx}{1\ !}+\frac{{{(bx)}^{2}}}{2\ !}+….. } $ The coefficient of $ x^{r}=e^{a}.\frac{b^{r}}{r\ !} $ .
BETA
