Sequence And Series Question 165
Question: If $ t_{n}=\frac{1}{4}(n+2),(n+3) $ for n = 1, 2, 3,?? then $ \frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+….+\frac{1}{t_{2003}}= $
[EAMCET 2003]
Options:
A) $ \frac{4006}{3006} $
B) $ \frac{4003}{3007} $
C) $ \frac{4006}{3008} $
D) $ \frac{4006}{3009} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ t_{n}=\frac{1}{4}(n+2)(n+3) $ , then $ \frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+…..+\frac{1}{t_{2003}} $ $ =4[ \frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+….+\frac{1}{(2005).(2006)} ] $ $ =4[ \frac{1}{3}-\frac{1}{2006} ] $ $ =4.\frac{2003,}{3(2006)}=\frac{4006}{3009} $ .
BETA
