Sequence & Series Nth Term Of Special Series Sum To N Terms And Infinite Number Of Terms Question 38
Question: The value of $ \frac{1}{(1+a)(2+a)}+\frac{1}{(2+a)(3+a)} $ $ \frac{1}{(3+a)(4+a)} $ + ….. + $ \infty $ is, (where a is a constant) [AMU 2005]
Options:
A) $ \frac{1}{1+a} $
B) $ \frac{2}{1+a} $
C) $ \infty $
D) None of these
Answer:
Correct Answer: A
Solution:
$ \frac{1}{(1+a)(2+a)} $ $ +\frac{1}{(2+a)(3+a)}+\frac{1}{(3+a)(4+a)}+…..+\infty $ n th term of series $ T_{n} $ $ =\frac{1}{(n+a)(n+1+a)} $ $ =\frac{1}{n+a}-\frac{1}{n+1+a} $ $ T_1=\frac{1}{1+a}-\frac{1}{2+a} $ ; $ T_2=\frac{1}{2+a}-\frac{1}{3+a} $ , $ T_3=\frac{1}{3+a}-\frac{1}{4+a} $ ………………… ………………… $ {T_{n-1}}=\frac{1}{n-1+a}-\frac{1}{n+a} $ , $ T_{n}=\frac{1}{n+a}-\frac{1}{n+1+a} $ \ $ S_{n}=T_1+T_2+T_3+……+T_{n} $ $ =\frac{1}{1+a}-\frac{1}{n+1+a} $ $ =\frac{n}{(1+a)(n+1+a)} $ $ S_{n}=\frac{1}{(1+a)( 1+\frac{1}{n}+\frac{a}{n} )} $ $ {S_{\infty }}=S_{n} $ , when $ n\to \infty $ \ $ {S_{\infty }} $ $ =\frac{1}{(1+a)} $ .
BETA
