Why are these two expressions different in this induction problem? 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.

Elevating comment to answer, at suggestion of OP: 
In the last term of the last displayed equation, there is a $5+1$ where there should be a $5+2$. Jonas Meyer notes that this correction gets rid of the discrepancy. 
