Search any question & find its solution
Question:
Answered & Verified by Expert
If $t_n=\frac{1}{4}(n+2)(n+3)$ for $n=1,2,3 \ldots$, then $\frac{1}{t_1}+\frac{1}{t_2}+\ldots+\frac{1}{t_{2003}}$ is equal to
Options:
Solution:
2058 Upvotes
Verified Answer
The correct answer is:
$\frac{4006}{3009}$
We have,
$t_n=\frac{1}{4}(n+2)(n+3)$
$\frac{1}{t_n}=\frac{4}{(n+2)(n+3)}$
$=4\left[\frac{1}{n+2}-\frac{1}{n+3}\right]$
Put $n=1,2,3, \ldots, 2003$
$\frac{1}{t_1}=4\left(\frac{1}{3}-\frac{1}{4}\right)$
$\begin{aligned} & \frac{1}{t_2}=4\left(\frac{1}{4}-\frac{1}{5}\right) \\ & \vdots \quad \vdots \quad \vdots\end{aligned}$
$\frac{1}{t_{2003}}=4\left(\frac{1}{2005}-\frac{1}{2006}\right)$
On adding these terms, we get
$\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\ldots+\frac{1}{t_{2003}}=4\left[\frac{1}{3}-\frac{1}{2006}\right]$
$=4\left[\frac{2006-3}{3 \cdot 2006}\right]=\frac{2 \cdot 2003}{3 \cdot 1003}=\frac{4006}{3009}$
$t_n=\frac{1}{4}(n+2)(n+3)$
$\frac{1}{t_n}=\frac{4}{(n+2)(n+3)}$
$=4\left[\frac{1}{n+2}-\frac{1}{n+3}\right]$
Put $n=1,2,3, \ldots, 2003$
$\frac{1}{t_1}=4\left(\frac{1}{3}-\frac{1}{4}\right)$
$\begin{aligned} & \frac{1}{t_2}=4\left(\frac{1}{4}-\frac{1}{5}\right) \\ & \vdots \quad \vdots \quad \vdots\end{aligned}$
$\frac{1}{t_{2003}}=4\left(\frac{1}{2005}-\frac{1}{2006}\right)$
On adding these terms, we get
$\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\ldots+\frac{1}{t_{2003}}=4\left[\frac{1}{3}-\frac{1}{2006}\right]$
$=4\left[\frac{2006-3}{3 \cdot 2006}\right]=\frac{2 \cdot 2003}{3 \cdot 1003}=\frac{4006}{3009}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.