Sequence And Series Question 636
Question: $ 1+\frac{a-b}{a}+\frac{1}{2,!}{{( \frac{a-b}{a} )}^{2}}+\frac{1}{3,!}{{( \frac{a-b}{a} )}^{3}}+……\infty = $
Options:
A) $ x= $
B) $ e^{a} $
C) $ \frac{e}{{e^{b/a}}} $
D) $ \frac{e}{{e^{a/b}}} $
Show Answer
Answer:
Correct Answer: C
Solution:
$ 1+\frac{a-b}{a}+\frac{1}{2\ !}{{( \frac{a-b}{a} )}^{2}}\ +\frac{1}{3\ !}{{( \frac{a-b}{a} )}^{3}}+…….. $ $ =1+\frac{( \frac{a-b}{a} )}{1\ !}+\frac{{{( \frac{a-b}{a} )}^{2}}}{2\ !}+\frac{{{( \frac{a-b}{a} )}^{3}}}{3\ !}+……. $ $ ={e^{(a-b)/a}}={e^{1-(b/a)}}=e^{1}.{e^{-b/a}}=\frac{e}{{e^{b/a}}} $ .
BETA
