Mathematical Induction: sum[k=1,n][1/((2k-1)(2k+1)] = n/(2n+1), for n>= 1

[simcan18]

Can someone help with the following in the attachment?

. . . . .$\displaystyle \displaystyle \sum_{k=1}^n\, \dfrac{1}{(2k\, -\, 1)\, (2k\, +\, 1)}\, =\, \dfrac{n}{2n\, +\, 1}\, \forall\, n\, \geq\, 1$

 

[Deleted member 4993]

Can someone help with the following in the attachment?

. . . . .$\displaystyle \displaystyle \sum_{k=1}^n\, \dfrac{1}{(2k\, -\, 1)\, (2k\, +\, 1)}\, =\, \dfrac{n}{2n\, +\, 1}\, \forall\, n\, \geq\, 1$

Please share your work/thoughts with us.

 

[simcan18]

Mathematical Induction

I'm stuck on how to get started..do I use a base case of n=1??

 

[Deleted member 4993]

I'm stuck on how to get started..do I use a base case of n=1??

Yes...