Question

Please provide complete solution.​

Answer 1

Answer:

S=∑$\frac{2}{n*(n+2)}=$ $\frac{3}{2} -(\frac{1}{n+1} +\frac{1}{n+2})$

Step-by-step explanation:

using partial fraction:

$\frac{2}{n*(n+2)}=\frac{A}{n} + \frac{B}{n+2}$

$2=A(n+2)+B(n)\\2=2A+n(A+B)$

⇒$A=1 and A+B=0$

$B=-A=-1$

therefore

$\frac{2}{n*(n+2)}=\frac{1}{n} - \frac{1}{n+2}$

suppose n=4

then expanding each term gives

∑$\frac{2}{n*(n+2)}$ =∑$(\frac{1}{n} - \frac{1}{n+2})$

             =$(1-\frac{1}{3}) +(\frac{1}{2} -\frac{1}{4}) +(\frac{1}{3} -\frac{1}{5} ) +(\frac{1}{4} -\frac{1}{6} )$

notice that n=1 term have 1 and -1/3

Also n=3 term have 1/3 and 1/5

this term 1/3 gets cancel out in n=1 and n=3

Similarly 1/4 also get cancel out in n=2 and n=4.

$=1+ \frac{1}{2} -\frac{1}{5} -\frac{1}{6}$

$=\frac{3}{2} -\frac{1}{5} -\frac{1}{6}$

now when n=4 we have

$=\frac{3}{2} -\frac{1}{4+1} -\frac{1}{4+2}$

therefore for n term we will have

$=\frac{3}{2} -\frac{1}{n+1} - \frac{1}{n+2}$

$=\frac{3}{2} -(\frac{1}{n+1} + \frac{1}{n+2})$