Question

lf $t_n=\frac{1}{4}(n+2)(n+3)$ for $n=1,2,3$.
then $\frac{1}{t_I}+\frac{1}{t_2}+\frac{1}{t_3}+\dots \dots ..+\frac{1}{t_{20}}=$

• A. $\frac{4006}{3006}$
• B. $\frac{4003}{3007}$
• C. $\frac{4006}{3008}$
• D. $\frac{80}{69}$

Answer 1

The correct option is D $\frac{80}{69}$

If $t_n=\frac{1}{4}(n+2)(n+3)$ for $n=1,2,3$

Then ,

$=\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+.......+\frac{1}{t_{20}}$

Now,
$t_1=\frac{1}{4}(1+2)(1+3)$

$t_1=\frac{1}{4}\left(3\right)\left(4\right)$
$t_2=\frac{1}{4}(2+2)(2+3)$

$t_2=\frac{1}{4}\left(4\right)\left(5\right)$

$t_3=\frac{1}{4}(3+2)(3+3)$

$t_3=\frac{1}{4}\left(5\right)\left(6\right)$
.
.
.
.
$t_{20}=\frac{1}{4}(20+2)(20+3)$

$t_{20}=\frac{1}{4}\left(22\right)\left(23\right)$
Now,
$=\frac{1}{t_1}+\frac{1}{t_2}+\frac{1}{t_3}+.......+\frac{1}{t_{20}}$

$=\frac{4}{3\times 4}+\frac{4}{4\times 5}+...........\frac{4}{22\times 23}$

$=4[\frac{1}{3\times 4}+\frac{1}{4\times 5}+........\frac{1}{22\times 23}]$

$=4[\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+........-\frac{1}{23}]$

$=4[\frac{1}{3}-\frac{1}{23}]$

$=4\left[\frac{23-3}{69}\right]$

$=\left[\frac{80}{69}\right]$