Question

The sum to $20$ terms of $\frac{3}{1.2}\left(\frac{1}{2}\right)+\frac{4}{2.3}{\left(\frac{1}{2}\right)}^2+\frac{5}{3.4}{\left(\frac{1}{2}\right)}^3+...$, is

• A. $1-\frac{1}{{21.2}^{20}}$
• B. $1-\frac{1}{{20.2}^{21}}$
• C. $1-\frac{1}{{19.2}^{20}}$
• D. None of these

Answer 1

Answer: (A)

$1-\frac{1}{{21.2}^{20}}$

The general term:
$t_r=\frac{r+2}{r(r+1)}.{\left(\frac{1}{2}\right)}^r=\frac{2(r+1)-r}{r(r+1)}.{\left(\frac{1}{2}\right)}^r=(\frac{2}{r}-\frac{1}{r+1}).{\left(\frac{1}{2}\right)}^r=\frac{1}{r}.{\left(\frac{1}{2}\right)}^{r-1}-\frac{1}{r+1}.{\left(\frac{1}{2}\right)}^r$
Thus, we have the series:
$r=1:\frac{1}{1}.{\left(\frac{1}{2}\right)}^{1-1}-\frac{1}{1+1}.{\left(\frac{1}{2}\right)}^1=1-\frac{1}{2^2}$
$r=2:\frac{1}{2}.{\left(\frac{1}{2}\right)}^{2-1}-\frac{1}{2+1}.{\left(\frac{1}{2}\right)}^2=\frac{1}{2^2}-\frac{1}{3.2^2}$
$r=3:\frac{1}{3}.{\left(\frac{1}{2}\right)}^{3-1}-\frac{1}{3+1}.{\left(\frac{1}{2}\right)}^3=\frac{1}{{3.2}^2}-\frac{1}{4.2^3}$
...
$r=20:\frac{1}{20}.{\left(\frac{1}{2}\right)}^{20-1}-\frac{1}{20+1}.{\left(\frac{1}{2}\right)}^{20}=\frac{1}{{20.2}^{19}}-\frac{1}{21.2^{20}}$
On adding the above terms, we have:
$S=1-\frac{1}{21.2^{20}}$
Hence, (a) is correct.