Question

Sum to $n$ terms of the series $\frac{1}{\mathrm{1.2.3}}+\frac{3}{\mathrm{2.3.4}}+\frac{5}{\mathrm{3.4.5}}+\frac{7}{\mathrm{4.5.6}}+....$is

• A. $\frac{n(n+1)}{2(n+2)(n+3)}$
• B. $\frac{n(3n+5)}{4(n+1)(n+2)}$
• C. $\frac{1}{6}-\frac{5}{(n+1)(n+4)}$
• D. $\frac{n(3n+1)}{4(n+1)(n+2)}$

Answer 1

The correct option is D $\frac{n(3n+1)}{4(n+1)(n+2)}$

$\frac{1}{\mathrm{1.2.3}}+\frac{3}{\mathrm{2.3.4}}+\frac{5}{\mathrm{3.4.5}}+\frac{7}{\mathrm{4.5.6}}+....uptonterms$
$=\sum _{r=1}^n\frac{2r-1}{r(r+1)(r+2)}$
$=\sum _{r=1}^n\frac{2}{(r+1)(r+2)}-\frac{1}{r(r+1)(r+2)}$
$=2\sum _{r=1}^n\frac{(r+2)-(r+1)}{(r+1)(r+2)}-\frac{1}{2}(\sum _{r=1}^n\frac{1}{r(r+1)}-\frac{1}{(r+1)(r+2)})$
$=2\sum _{r=1}^n(\frac{1}{r+1}-\frac{1}{r+2})-\frac{1}{2}\sum _{r=1}^n(\frac{1}{r}-\frac{1}{r+1})+\frac{1}{2}\sum _{r=1}^n(\frac{1}{r+1}-\frac{1}{r+2})$
$=2(\frac{1}{2}-\frac{1}{n+2})-\frac{1}{2}(1-\frac{1}{n+1})+\frac{1}{2}(\frac{1}{2}-\frac{1}{n+2})$
$=\frac{n}{n+2}-\frac{n}{2(n+1)}+\frac{n}{4(n+2)}$
After simplifying, we get
$=\frac{n(3n+1)}{4(n+1)(n+2)}$
Hence, option 'D' is correct.