I'm not sure how my professor got from: $$n(n+1)(2n+1)+6(n+1)^2 ~\text{to}~ (n+1)(n(2n+1)+6(n+1))$$
I have a feeling it's an algebra thing but I just can't seem to get it. Would really appreciate the help.
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the_candyman
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Bill Ngo
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6Your teacher took out the common factor $(n+1)$ from $n(n+1)(2n+1)$ to leave $n(2n+1)$ and from $6(n+1)^2$ to leave $6(n+1)$. For rings (including real numbers) you have distributivity: $xy+xz=x(y+z)$ – Henry Oct 27 '18 at 18:40
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Take (n+1) commonly out side. – Avinash N Oct 27 '18 at 18:40
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Do you mean $$\frac{n(n+1)(2n+1)}{6}+(n+1)^2$$? – Dr. Sonnhard Graubner Oct 27 '18 at 18:42
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@the_candyman. I don't think the OP meant "proof". I think the OP mean "prof" and is in "my prof" = "my professor". – fleablood Oct 27 '18 at 18:55
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@fleablood lol, I fixed it. Thanks. – the_candyman Oct 28 '18 at 18:55
3 Answers
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You can expand your equation to :
$$n(n+1)(2n+1)+6(n+1)^2 = n(n+1)(2n+1)+6(n+1)(n+1)$$
On the right hand side you will notice that $n+1$ occurs in each group of terms and this can be factored out as follows:
$$n(n+1)(2n+1)+6(n+1)(n+1) = (n+1)\cdot\left[n(2n+1)+6(n+1)\right]$$
user32882
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It is an algebra thing.
$\color{blue}{n}\color{green}{(n+1)}\color{blue}{(2n+1)}$ and $\color{red}6\color{green}{(n+1)}^{\color{red}2}$ both have $\color{green}{(n+1)}$ as a common factor. So the common factor can be factored out:
$\color{blue}{n}\color{green}{(n+1)}\color{blue}{(2n+1)} + \color{red}6\color{green}{(n+1)}^{\color{red}2} = \color{green}{(n+1)}[\color{blue}{n}\color{blue}{(2n+1)} + \color{red}6\color{green}{(n+1)}^{\color{red}1}]$
fleablood
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One way is to start with $$n(n+1)(2n+1)+6(n+1)^2$$
and factor $(n+1)$ to get $$ (n+1)(n(2n+1)+6(n+1))$$
The other way which is lengthy is to multiply both sides and see if you get the same expressions. ( not recommended)
Mohammad Riazi-Kermani
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