simplifying formula $\frac{(k-1)(k-1 +1) (2(k-1)+1)}6$ I'm trying to understand my textbook's example (it's part of a mathematical induction problem) and I just don't get how the author managed to simplify this formula:
$$\frac{(k-1)(k-1 +1) (2(k-1)+1)}6    =     \frac{(k-1) k(2k - 1)}6$$
Could anyone explain in simple steps how they got from the left hand to the right hand? My math skills are very basic.
Thanks
 A: He simply carried out the calculations $k-1+1=k$ and $2(k-1)+1=(2k-2)+1=2k-1$; the $k-1$ in the numerator and the $6$ in the denominator remained unchanged.
A: We can simplify the numerator in two steps. First consider the term $k-1+1$. This is clearly $k$ since the $1$s cancel. For the term $2(k-1)+1$ first distribute the 2 by multiplying both $k$ and $-1$ by 2. Now we have the expression
$$
2(k-1)+1=2k-2+1=2k-1
$$
Now just plug these simplified bits back in to the original problem to get
$$
(k-1)(k+1-1)(2(k-1)+1)=(k-1)k(2k-1)
$$
A: Leaning how to deduce missing steps in inferences is an essential skill required to follow proofs. Let's consider your example. It is an equation between two products. To verify such an equation we can first cancel any obvious nonzero common factors from both sides. In your equation one immediately recognizes the common factors $\rm\:k-1\:$ and $\:1/6,\:$ so cancelling them leaves us with
$$ (k-1 +1)\: (2(k-1)+1)\   =\    k\:(2k - 1) $$
Now let's simplify the first term on the LHS: $\: k-1+1 = k,\:$ which matches the first term on the RHS. Cancelling this common factor $\:k\:$ leaves us only to verify that
$$ 2(k-1)+1  =\   2k - 1 $$
Now, subtracting $1$ from both sides leaves us with
$$ 2(k-1)  =\   2k - 2 $$
which is the hairiest law applied, the distributive law $\ a(b+c) = ab + ac.\:$ Note how you might have discovered the distributive law this way, even if you were not aware of it, or if you had forgotten about it. By stripping away all the simpler parts of the derivation we have managed to isolate the only complicated part of the derivation. (This can  prove especially useful when you are working with functions satisfying unfamiliar exotic identities. It will allow you to reduce the proof to (an instance of) such an identity, whose proof you can then probably easily look up).
Now it is clear how to apply the arithmetic laws to derive the RHS from the LHS.
Analogous techniques work generally to transform hairy equations into simpler equivalent forms, by successively whittling away simpler stuff, until it becomes clear precisely how the laws and identities were applied in more hairy parts of the derivation. 
Note $ $ See also this answer on extraneous solutions arising from non $1\!-\!1$ transformations, such as squaring both sides of an equation. 
