# 1+3/4+5/36+7/144+...+2n+1/n^2(n+1)^2=2-1/(n+1)^2 prove using mathematical induction

pranay1983 | Student

For n =1,

1 + 3/4 = 2 - 1/4

Thus result is true for n = 1,

We assume it is true for n = k,

1+3/4+5/36+7/144+...+(2k+1)/(k+1)^2 = 2-1/(k+1)^2

now for n = k+1

1+3/4+5/36+7/144+...+(2k+1)/(k+1)^2 + (2k+3)/(k+2)^2 = 2-1/(k+1)^2 + (2k+3)/(k+2)^2 = 2 - 1/(k+2)^2

aruv | Student

`1+3/4+5/36+7/144+...........+(2n+1)/(n^2(n+1)^2)=2-1/(n+1)^2`

Let define P(n) as

`P(n)=3/4+5/36+7/144+.......+(2n+1)/(n^2(n+1)^2)=2-1-1/(n+1)^2=1-1/(n+1)^2`

Let n=1

`P(1): 3/4=1-1/2^2=3/4`    ,which is true.

Let n=k is true i.e.

`P(k)=3/4+5/36+7/144+.....+(2k+1)/(k^2(k+1)^2)=1-1/(k+1)^2`

Now we wish to prove that n=k+1 is true when n=k is true.

`p(k+1)=3/4+5/36+......+(2k+1)/(k^2(k+1)^2)+(2(k+1)+1)/((k+1)^2(k+1+1)^2)=1-1/(k+1+1)^2`

`LHS=3/4+5/36+...+(2k+1)/(k^2(k+1)^2)+(2(k+1)+1)/((k+1)^2(k+1+1)^2)`

=`1-1/(k+1)^2+(2(k+1)+1)/((k+1)^2(k+1+1)^2)`    since P(k) is true.

`=1-(1/(k+1)^2)(1-(2k+3)/(k+2)^2)`

`=1-1/(k+1)^2(k^2+4+4k-2k-3)/(k+2)^2`

`=1-1/(k+1)^2 (k+1)^2/(k+2)^2`

`=1-1/(k+2)^2`

`=1-1/(k+1+1)^2=RHS`

Thus P(k+1) is true when P(k) is true.

Thus P(n) is true for n ,n is natural number.

Hence proved.

`P(2):3/4 =2-1/(2)^2`

`7/4=`