I have two questions that tie to together how many number of terms is there when i factoring out $k + 1$ when $\frac{k(k + 1)(2k + 1)}{6} + \frac{6(x + 1)^2}{6}$ and how come the common factor is $x + 1$ in the first place and if $k(k + 1)$ and $(2k + 1)$ and $\frac{6(x + 1)^2}{6}$ count up to be $3$ terms or am i mistaken how many number of terms do i have in this proof. i tried to factor out $k + 1$ out of all the terms but i end up with $k(k+1)$ i dont know how im really trying to understand proof by induction i understand the simpler ones but im stuck on this particular problem.


this site leads to the question im stuck on i see the steps but i thought factoring is dividing each term by the common factor or am i wrong

  • $\begingroup$ Are those $x$'s meant to be $k$'s, or are there two variables in this formula? What's our context here? $\endgroup$ Sep 13 '17 at 19:40
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    $\begingroup$ sorry i edited i saw the mistake i fixed it $\endgroup$ Sep 13 '17 at 19:45
  • $\begingroup$ \begin{eqnarray*} \frac{k \color{orange}{(k+1)}(2k+1)}{\color{orange}{6}}+\frac{6 \color{orange}{(k+1)}^2}{\color{orange}{6}} =\color{orange}{\frac{(k+1)}{6}} \left( k(2k+1)+6(k+1) \right) \end{eqnarray*} $\endgroup$ Sep 13 '17 at 19:47
  • $\begingroup$ how many terms do i have to factor out k + 1 $\endgroup$ Sep 13 '17 at 19:49
  • $\begingroup$ i understand what you did but still a little bit confusing to me $\endgroup$ Sep 13 '17 at 19:55

$$\begin{align} \frac{k(k+1)(2k+1)}{6}+\frac{6(k+1)^2}{6} &= \frac{k(k+1)(2k+1)+6(k+1)^2}{6}\\ &=\frac{kBC+6B^2}{6}\\ &= \frac{B(kC+6B)}{6} \end{align}$$

There are two terms present, before the $B$ is factored out. The first term has three factors, namely $k$, and $B=(k+1)$, and $C=(2k+1)$, but they all are multiplied together, so they make up one term.

Does that help?

  • $\begingroup$ yes i get i now thanks $\endgroup$ Sep 13 '17 at 20:20

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