<s>[INST] Proving $\sum_{i=1}^n (1-\frac{1}{(i+1)^2}) = \frac{n+2}{2n+2}$ using induction. My textbook has the following question:

Prove the follwing statement using induction for all natural numbers $n$
$$(1- \frac{1}{4})+(1- \frac{1}{9})+.......+(1- \frac{1}{(n+1)^2})=\frac{n+2}{2n+2}$$

So, I check both the sides for $n=1$. In that case LHS=RHS=$\frac{3}{4}$.
Now  I assume the statement to be true for $n=k$ which gives
$$
\sum_{i=1}^k (1-\frac{1}{(i+1)^2}) = \frac{k+2}{2k+2}
$$
Now I evaluate the original statement for $n=k+1$ which leaves us with the
$$ LHS= \frac{3k^3+16k^2+26k+14}{2(k+1)(k+2)^2} $$
And we have to prove this LHS to be equal to RHS which is
$$
RHS= \frac{k+3}{2k+4}
$$
But these (new) LHS and RHS don't seem to be equal. And hence I am not able to complete the proof.
How should I proceed?
A solution without induction is also welcome.

Book: Comprehensive Algebra VOL-1
Author: Vinay Kumar
Publisher: McGraw Hill Education.
 [/INST] For $n=2$ we have
$$\sum_{i=1}^{2}(1-\frac{1}{(i+1)^{2}})=(1-\frac{1}{4})+(1-\frac{1}{9})=\frac{59}{36}$$ but $$\frac{n+2}{2n+2}\big|_{n=2}=\frac{2+2}{2(2)+2}=\frac{2}{3}\neq\frac{59}{36}.$$
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