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


Let define P(n) as


Let n=1

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

Let n=k is true i.e.


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



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



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`