Binomial Theorem And Its Simple Applications Question 188
Question: If $ a_{k}=\frac{1}{k(k+1)}, $ for $ k=1,,2,,3,,4,…..,,n $ , then $ {{( \sum\limits_{k=1}^{n}{a_{k}} )}^{2}}= $
[EAMCET 2000]
Options:
A) $ ( \frac{n}{n+1} ) $
B) $ {{( \frac{n}{n+1} )}^{2}} $
C) $ {{( \frac{n}{n+1} )}^{4}} $
D) $ {{( \frac{n}{n+1} )}^{6}} $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \sum\limits_{k=1}^{n}{a_{k}}=\sum\limits_{k=1}^{n}{\frac{1}{k,(k+1)}} $ = $ ( 1-\frac{1}{2} )+( \frac{1}{2}-\frac{1}{3} )+…+( \frac{1}{n}-\frac{1}{n+1} ) $ = $ 1-\frac{1}{n+1}=\frac{n}{n+1} $
$ {{( \sum\limits_{k=1}^{n}{a_{k}} )}^{2}}={{( \frac{n}{n+1} )}^{2}} $ .
BETA
