Sequence And Series Question 554
Question: The sum of the series $ 1+\frac{3}{2,!}+\frac{7}{3,!}+\frac{15}{4,!}+…..to,\infty $ is
[Kerala (Engg.) 2005]
Options:
A) $ e(e+1) $
B) $ e,(1-e) $
C) $ 3e-1 $
D) $ 3e $
E) (e) $ e,(e-1) $
Show Answer
Answer:
Correct Answer: E
Solution:
(e) $ 1+\frac{3}{2,!}+\frac{7}{3,!}+\frac{15}{4,!}+…. $ $ =,(1-1)+( \frac{2}{1!}-\frac{1}{1!} )+( \frac{2^{2}}{2,!}-\frac{1^{2}}{2,!} )+( \frac{2^{3}}{3,!}-\frac{1^{3}}{3,!} )+… $ $ =( 1+\frac{2}{1!}+\frac{2^{2}}{2!}+\frac{2^{3}}{3!}+…. ) $ $ -( 1+\frac{1}{1!}+\frac{1^{2}}{2!}+\frac{1^{3}}{3!}+… ) $ $ =e^{2}-e $ = $ e(e-1) $ .
BETA
