Latest update

simple Algebraic Manipulation (factoring?)

2018-03-12 20:23:32

I got stuck on some simple algebraic manipulation in a mathematical induction problem.

The statement I have is

$$\frac{k(k + 1)(2k + 1)}{6} +(k + 1)^2,$$

and I know it is equal to

$$\frac{k+1}{6}(k(2k + 1) + 6(k + 1)),$$

but I don't know the steps from one to the other, if somebody could please clarify for me.

$$\dfrac {k(k+1)(2k+1)}6+(k+1)^2=\dfrac {k(k+1)(2k+1)}6+\dfrac{6(k+1)^2}{6}=\dfrac{(k+1)[k(2k+1)+6(k+1)]}{6}$$

After factoring out $k+1$. Can you figure out the rest?

$\frac{k(k + 1)(2k + 1)}{6} +(k + 1)^2$

RESIST the urge to expand. That's a common mistake. Factoring is recognizing "chunks" and pulling them out. So there isn't any reason to "blend" before you pull out the chunks. You may have to "blend" to put the chunks in different places but you should put that off as long as you can.

So first things first. Put over common denominator.

That would be $6$.

$\frac{k(k + 1)(2k + 1)}{6} +(k + 1)^2=$

$\frac {k(k+1)(2k+1)}6 + \frac {6(k+1) • $$\dfrac {k(k+1)(2k+1)}6+(k+1)^2=\dfrac {k(k+1)(2k+1)}6+\dfrac{6(k+1)^2}{6}=\dfrac{(k+1)[k(2k+1)+6(k+1)]}{6}$$ After factoring out$k+1$. Can you figure out the rest? 2018-03-12 21:16:39 •$\frac{k(k + 1)(2k + 1)}{6} +(k + 1)^2$RESIST the urge to expand. That's a common mistake. Factoring is recognizing "chunks" and pulling them out. So there isn't any reason to "blend" before you pull out the chunks. You may have to "blend" to put the chunks in different places but you should put that off as long as you can. So first things first. Put over common denominator. That would be$6$.$\frac{k(k + 1)(2k + 1)}{6} +(k + 1)^2=\frac {k(k+1)(2k+1)}6 + \frac {6(k+1)^2}6=\frac {k(k+1)(2k+1) + 6(k+1)^2}6$. Now look for common "chunks". There is a$(k+1)$that both terms in the numerator have. Pull it out.$\frac {(k+1)[k(2k+1) + 6(k+1)]}6\$

2018-03-12 21:22:23