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.
 A: There is a flaw in the logic in this posting:
begin quote:
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.
end of quote
This is not valid reasoning.  You're saying "Let's prove $A$, as follows: If $A$ then $3/4 = 3/4$, which is true."  Very well then, I shall prove that Socrates was John F. Kennedy: If Socrates was John F. Kennedy, then Socrates died in 1963, and hence, Socrates was mortal.  Which is true."  To say "If B then A. And A is true.  Therefore B." is a standard erroneous form of reasoning.  Writers on logic since the time of Aristotle have identified this as a fallacy.  A correct way of reasoning is this:
$$
\frac{1+2}{1(1+1)2^1} = \frac34 = 1 - \frac14 = 1-\frac{1}{(1+1)2^1}.
$$
Therefore the proposition in case $n=1$.
And then you repeat the same fallacy in your inductive step.
You should write "$=$" between things ALREADY KNOWN TO BE EQUAL.  Then when you write
$$
A = B = C = \cdots = Z,
$$
then that proves that $A=Z$.  You should not write
$$
A = Z
$$
therefore
$$
B=Y
$$
therefore
$$
C=X
$$
etc. etc. etc. etc.
Therefore
$$
3=3
$$
which is true.  Therefore $A=Z$.
A: 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. 
