Standard induction proof of a sum

Question

Via Induction, Prove

$$ \sum_{i=2}^n\frac{1}{i^2-1} = \frac{(n-1)(3n+2)}{4n(n+1)}$$

Answer 1

If you need to prove $\sum_{i=1}^n a_i = f(n)$ by induction the key point is prove $f(n) + a_{n+1} = f(n+1)$.

So all you need to do is prove that $\frac{(n-1)(3n+2)}{4n(n+1)} + \frac 1{(n+1)^2 - 1} = \frac {n(3(n+1)+2)}{4(n+1)(n+2)}$. If you can prove that equation you are done.

As $(n+1)^2 - 1 = n^2 + 2n = n(n+2)$ putting over a common denominator look quite doable.

$\frac{(n-1)(3n+2)}{4n(n+1)} + \frac 1{(n+1)^2 - 1}=$

$\frac {(n-1)(3n+2)}{4n(n+1)}\cdot \frac {n+2}{n+2} + \frac 1{n(n+2)}\cdot \frac {4(n+1)}{4(n+1)}=$

$\frac {(n-1)(3n+2)(n+2) +4(n+1)}{4n(n+1)(n+2)}$

Okay, we want this to equal $\frac {n(3(n+1)+2)}{4(n+1)(n+2)}=\frac {n^2(3n+5)}{4n(n+1)(n+2)}$

So we just need to show $(n-1)(3n+2)(n+2) +4(n+1)= n^2(3n+5)$ and Thats straightforward:

$(n-1)(3n+2)(n+2) +4(n+1) = 3n^3 + (6+2-3)n^2 + (4-6-2+4)n + (-4 + 4)= 3n^2 + 5n^2= n^2(3n+5)$. We are done.

.... once we do the base case of $n= 2$

We have to prove $\frac 1{2^2 -1} = \frac {(2-1)(3\cdot 2 +2)}{4\cdot 2\cdot (2+1)}$.

LHS is $\frac 13$ and RHS is $\frac 8{24}= \frac 13$.

We are done.