Prove with $n \ge 1$:
$\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot4}+\cdots+\frac{n+2}{n(n+1)2^n} = 1 - \frac{1}{(n+1)2^n}$
First, I prove it for $n=1$: $\left(\frac{1+2}{1(1+1)2^1} = 1-\frac{1}{(1+1)2^1}\right) \implies \left(\frac{3}{4} = 1- \frac{1}{4}\right) \implies \left(\frac{3}{4} = \frac{3}{4}\right)$ Which is true.
So I will now assume this: $\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot2}+\cdots+\frac{n+2}{n(n+1)2^n} = 1 - \frac{1}{(n+1)2^n}$
And I want to prove it for $n+1$, i.e:
$\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot2}+\cdots+\frac{n+2}{n(n+1)2^n} + \frac{n+3}{(n+1)(n+2)2^{n+1}} = 1 - \frac{1}{((n+1)+1)2^{n+1}}$
This is how I tried to prove it:
$\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot2}+\cdots+\frac{n+2}{n(n+1)2^n} + \frac{n+3}{(n+1)(n+2)2^{n+1}} =$
$ 1 - \frac{1}{(n+1)2^n} + \frac{n+3}{(n+1)(n+2)2^{n+1}}$
Before continuing, I usually like to grab a calculator, give a value to $n$ and evaluate my current expression with the expression I want to reach. If both values are equal, it means I'm doing okay.
So I took $n=5$ and evaluated the expression I want to reach:
$1 - \frac{1}{(5+2)\cdot2^5+1} = \frac{447}{448}$ Then, still with $n = 5$ I evaluated my current expression: $1 - \frac{1}{(5+1)\cdot2^5}+\frac{5+3}{(5+1)(5+1)\cdot2^{5+1}} = \frac{575}{576}$
So I got $\frac{447}{448}$ for the expression I want to reach and $\frac{575}{576}$ for what I got so far. Something went wrong.
My problem with this is that I haven't done any calculations yet. All my steps so far were rather mechanical - things I always do with mathematical induction.
Maybe I simply evaluated them wrongly. But I can't see it - I've tested it many times already.
Why are both expressions different? They should be the same.