JEE maths#22

$S = \displaystyle\sum_{k=1}^{n} \dfrac{k+3}{k(k+1)(k+2)}$

$\Rightarrow \displaystyle\sum_{k=1}^{n} \dfrac2{k(k+1)}\cdot\dfrac{k+3}{2k+4}$

$\Rightarrow \displaystyle\sum_{k=1}^{n} \dfrac2{k(k+1)}\cdot\left(1 - \dfrac{k+1}{2k+4}\right)$

$\Rightarrow \displaystyle\sum_{k=1}^{n} \dfrac2{k(k+1)} - \dfrac1{k(k+2)}$

$\Rightarrow 2\left(1-\dfrac1{n+1}\right) - \dfrac12\left(1 + \dfrac12 - \dfrac1{n+1} - \dfrac1{n+2}\right)$

$\Rightarrow \dfrac54 - \dfrac3{2(n+1)}+ \dfrac1{2(n+2)}$

$\Rightarrow \dfrac54 - \dfrac12\left(\dfrac3{(n+1)} - \dfrac1{(n+2)}\right)$

$\Rightarrow \dfrac54 - \dfrac{(2n+5)}{2(n+1)(n+2)}$

$\therefore \boxed{K = \dfrac54}$

$\begin{aligned} S & = \sum_{k=1}^n \frac {k+3}{k(k+1)(k+2)} & \small \color{#3D99F6} \text{By partial fractions} \\ & = \frac 12 \sum_{k=1}^n \left(\frac 3k - \frac 4{k+1} + \frac 1{k+2} \right) \\ & = \frac 12 \sum_{k=1}^n \left(\frac 3k - \frac 3{k+1} - \frac 1{k+1} + \frac 1{k+2} \right) \\ & = \frac 12 \left(\frac 31 - \frac 3{n+1} - \frac 12 + \frac 1{n+2} \right) \\ & = \frac {5(n+1)(n+2)-6(n+2)+2(n+1)}{4(n+1)(n+2)} \\ & = \frac {5n^2+11n}{4(n+1)(n+2)} \\ & = \frac {5n^2+15n+10-4n-10}{4(n+1)(n+2)} \\ & = \frac {5(n+1)(n+2)-2(2n+5)}{4(n+1)(n+2)} \\ & = \frac 54 - \frac {2n+5}{2(n+1)(n+2)} \end{aligned}$

$\implies K = \boxed{\dfrac 54}$