Sequence And Series Question 29
Question: $ 1+\frac{a-bx}{1,!}+\frac{{{(a-bx)}^{2}}}{2,!}+\frac{{{(a-bx)}^{3}}}{3,!}+….\infty = $
Options:
A) $ {e^{a-bx}} $
B) $ {e^{a-bx}}-1 $
C) $ 1+a{\log_{e}}(a-bx) $
D) $ {e^{-bx}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ 1+\frac{(a-bx)}{1\ !}+\frac{{{(a-bx)}^{2}}}{2\ !}+\frac{{{(a-bx)}^{3}}}{3\ !}+……={e^{a-bx}} $ .
BETA
