Question

If ${\mathrm{t}}_{\mathrm{n}}=\left(\frac{1}{4}\right)\left(\mathrm{n}+2\right)\left(\mathrm{n}+3\right)$ for $\mathrm{n}=1,2,3...$then $\frac{1}{{\mathrm{t}}_1}+\frac{1}{{\mathrm{t}}_2}+\frac{1}{{\mathrm{t}}_3}+...\frac{1}{{\mathrm{t}}_{2003}}=?$

Answer 1

Explanation for the correct answer:

Finding the sum of the series:

Given that,

${\mathrm{t}}_{\mathrm{n}}=\left(\frac{1}{4}\right)\left(\mathrm{n}+2\right)\left(\mathrm{n}+3\right)$

Using summation,

$\begin{array}{rcl}\sum _{\mathrm{n}=1}^{\mathrm{n}=2003}\frac{1}{{\mathrm{t}}_{\mathrm{n}}}& =& \frac{1}{{\mathrm{t}}_1}+\frac{1}{{\mathrm{t}}_2}+...+\frac{1}{{\mathrm{t}}_{2003}}\\ & =& \sum _{\mathrm{n}=1}^{\mathrm{n}=2003}\frac{4}{\left(\mathrm{n}+2\right)\left(\mathrm{n}+3\right)}\\ & =& 4\sum _{\mathrm{n}=1}^{\mathrm{n}=2003}\left(\frac{1}{\mathrm{n}+2}-\frac{1}{\mathrm{n}+3}\right)\\ & =& 4\left(\frac{1}{3}-\frac{1}{2006}\right)\left[\because n=1\&2003\right]\\ & =& \frac{4006}{3009}\end{array}$

Hence, the sum of the series is$\begin{array}{l}\frac{\mathbf{4006}}{\mathbf{3009}}\end{array}$.