Distributing 2 times k equals k?

Question

in my current discrete mathmatics course I have this calculation at the end of a proof:

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

I dont under stand why 2*k ends up being k? I guessing there is some calculations being abstracted away, can someone please exaplain?

I think it would be clearer if you let $x=k+1$ and write the left hand as $\frac {kx+2x}2$. Then, all you are doing is factoring out the $x$ in the numerator. — lulu, Apr 22 '20 at 10:00
nice, but what is was looking for was more what happens inbetween the two, when you distribute 2, why does 2*k become k? — s3j80, Apr 22 '20 at 10:02
I don't understand. $kx+2x=(k+2)x$. That's just the distributive law. There is no sense in which $2k$ becomes $k$. — lulu, Apr 22 '20 at 10:05

score 1 · Answer 1 · answered Apr 22 '20 at 10:15

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With colors: $$\frac{\color{blue}{k}\color{red}{(k+1)}+\color{green}{2}\color{red}{(k+1)}}{2} =\frac{(\color{blue}{k}+\color{green}{2})\color{red}{(k+1)}}{2}$$ The red part $\color{red}{(k+1)}$ is seen in both terms of the left-hand-side, so can be "moved out" to the right.

answered Apr 22 '20 at 10:15

Jeppe Stig Nielsen

5,109
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so youre converting two (k + 1) to one, doesnt makr sense? – s3j80 Apr 22 '20 at 10:36
@s3j80 Do you (should you) know the distributive law, $(\color{blue}{A}+\color{green}{B})\color{red}{C} = \color{blue}{A}\color{red}{C} + \color{green}{B}\color{red}{C}$? – Jeppe Stig Nielsen Apr 22 '20 at 11:45
no! i had no idea! – s3j80 Apr 22 '20 at 15:08
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This is something very basic that you have to get familiar with before you follow a discrete mathematics course. – Jeppe Stig Nielsen Apr 22 '20 at 15:15
it just seems odd to me that the c that is to be distributed is behind the parantheses? – s3j80 Apr 22 '20 at 15:18
og jeg forstår ikke helt hvad det er du behandler som c'et i den her kontekst – s3j80 Apr 22 '20 at 15:21

score 0 · Answer 2 · answered Apr 22 '20 at 10:02

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Just factor out the $(k+1)$ term

or if you prefer expand and then factor:

$$\frac {k(k+1)+2(k+1)}2=\frac {k^2+3k+2}2=\frac {(k+1)(k+2)}2$$

answered Apr 22 '20 at 10:02

Henry

157,058

nice, exactly what i was looking for, but where does 3k come from? – s3j80 Apr 22 '20 at 10:04
@s3j80: $k(k+1)=k^2+k$ while $2(k+1)=2k+2$ so $k(k+1)+2(k+1)=k^2+k+2k+2=k^2+3k+2$ – Henry Apr 22 '20 at 10:05
Try with $k=10$ as an example: $\frac{10\times 11 +2\times 11}{2} = \frac{110+22}{2} =\frac{132}{2} = \frac{11\times 12}{2}$ – Henry Apr 22 '20 at 10:09
K * 1 provies k? – s3j80 Apr 22 '20 at 10:09
@s3j80: $k \times 1=k$ is certainly true – Henry Apr 22 '20 at 10:10
all right thank you, that clears up a lot of things – s3j80 Apr 22 '20 at 10:11
how does k^2+3k+2 , end up being (k+1)(k+2)? – s3j80 Apr 22 '20 at 10:17
@s3j80 Do you know how to expand $(k+1)(k+2)$? – Henry Apr 22 '20 at 11:28
hmm, i guess not, could you point me in the drection of link? – s3j80 Apr 22 '20 at 15:07
@s3j80 https://en.wikipedia.org/wiki/Distributive_property – Henry Apr 22 '20 at 15:33

Distributing 2 times k equals k?

2 Answers2