Q94 / 160 · TS EAMCET 2003 · Mathematics

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

  • A
    $\frac{4006}{3006}$
  • B
    $\frac{4003}{3007}$
  • C
    $\frac{4006}{3008}$
  • Correct answer
    D
    $\frac{4006}{3009}$

Solution

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 $\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+\ldots+\frac{1}{t_{2003}}=4\left[\frac{1}{3}-\frac{1}{2006}\right]$ $=4\left[\frac{2006-3}{3 \cdot 2006}\right]=\frac{2 \cdot 2003}{3 \cdot 1003}=\frac{4006}{3009}$

Hence, the correct answer is option: 'D'.

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