Mathematical induction problem: $\frac{1}{1\cdot3}+\frac{1}{2\cdot4}+\frac{1}{3\cdot5}+...+\frac{1}{n(n+2)}=\frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}$

Question

I have problem solving this induction. I must be missing something because i don't know how to do it. Can someone help? Thanks $$\frac{1}{1\cdot3}+\frac{1}{2\cdot4}+\frac{1}{3\cdot5}+...+\frac{1}{n(n+2)}=\frac{3}{4}-\frac{2n+3}{2(n+1)(n+2)}$$

Answer 1

HINT

Consider the arithmetic progression $(a_{n})_{n\in\mathbb{N}}$ where $a_{n}\neq 0$ for every $n\in\mathbb{N}$. Then one has that:

\begin{align*} \sum_{k=1}^{n}\frac{1}{a_{k}a_{k+2}} = \frac{1}{2r}\sum_{k=1}^{n}\frac{a_{k+2} - a_{k}}{a_{k}a_{k+2}} = \frac{1}{2r}\sum_{k=1}^{n}\left(\frac{1}{a_{k}} - \frac{1}{a_{k+2}}\right) \end{align*} where $r = a_{n + 1} - a_{n} \neq 0$ and $n\in\mathbb{N}$.

Can you take it from here?

Answer 2

Let's start with checking for $n=1$ $$\frac{1}{1×3}=\frac{3}{4}-\frac{5}{12}$$ So, it satisfies for $n=1$

Now, we prove that if it works for $k$ then it works for $k+1$

So, we need to prove $$\frac{3}{4}-\frac{2k+3}{2(k+1)(k+2)}+\frac{1}{(k+1)(k+3)}=\frac{3}{4}-\frac{2k+5}{2(k+2)(k+3)}$$ We cancel out 3/4 from both sides and take lcm to get

$$\frac{2(k+2)-(2k+3)(k+3)}{2(k+1)(k+2)(k+3)}=-\frac{2k+5}{2(k+2)(k+3)}$$ Cancelling some more stuff out we get

$$\frac{2(k+2)-(2k+3)(k+3)}{(k+1)}=-(2k+5)$$ Simplifying the numerator of LHS

$$\frac{-2k²-7k-5}{k+1}=-2k-5$$

Now , we can factor $-2k²-7k-5$ as $$-2k²-2k-5k-5=-2k(k+1)-5(k+1)=-(k+1)(2k+5)$$

Putting that in we can see that the LHS becomes the same as the RHS and hence the proof.

Note: All the cancellations I did were because none of them could be 0.