Proof by induction: conversion to equal expressions Lets suppose that I am trying to prove: $$1^2+...+n^2 = 1/6n(n+1)(2n+1)$$ via induction, but I am unable to convert (for sake of argument) $$1/6k(k+1)(2k+1)+(k+1)^2$$ to $$1/6(k+1)(k+2)(2k+3)$$ to prove that
$$1^2+...+k^2+(k+1)^2 = 1/6k(k+1)(2k+1) + (k+1)^2$$
so instead I manage to convert both of them to $$1/6(2n^3+9n^2+15n+6)$$
showing that they are equal. Would that be enough to consider this proof by induction complete are is my reasoning wrong?
 A: Your logic is backwards: you started with the conclusion and derived something true, rather than starting with what you know and deriving the conclusion.
For example, $0 = 1$ implies $0=0$ by multiplying both sides by zero, but that doesn't mean that $0=1$ is true!
If you want to use this line of reasoning, you need to argue that your steps are reversible. Then from the equation you derive, you can deduce that the statement you derived it from (the goal of the induction step) is true.
A: We need to prove the induction step, that is
$$1^2+...+k^2+(k+1)^2 =\frac{(k+1)(k+2)(2k+3)}6$$
and using the induction hypotesis, that is
$$1^2+...+k^2 =\frac{k(k+1)(2k+1)}6$$
you have properly derived
$$1^2+...+k^2+(k+1)^2 =\frac{k(k+1)(2k+1)}6+(k+1)^2=\frac{k(k+1)(2k+1)+6(k+1)^2}6$$
now we need a few algebra to obtain
$$k(k+1)(2k+1)+6(k+1)^2=(k+1)(2k^2+k+6k+6)=(k+1)(2k^2+7k+6)$$
and since 
$$(2k^2+7k+6)=(k+2)(2k+3)$$
we are done.
Your last step obtained converting all by the variable $n$ is not so much clear. You should complete the proof using the variable $k$ till the end to be more clear.
A: You have to show that
$$(n+1)^2+\frac{1}{6}n(n+1)(2n+1)=\frac{1}{6}(n+1)(n+2)(2n+3)$$
Can you show this?
Factorizing this we get $0$.
