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So I have the following problem

prove: Σ(n)(k+1) k+3 = n^2 / 2 + (7/2)n 
P(1) = 1 + 3 = 1^2 / 2 + 7 /2
P(1) = 4 = 4
So I assume it's true for n and attempt to prove it's true for n+1:
(n+1) + 3 = (n+1)^2 / 2 + (7/2)(n+1)
n+4 = (n^2 + 2n + 1) /2 + (7/2)(n+1)

and now I'm stuck. I have no idea how to make the functions equal. I can't really see a mistake either.

Jake Burns
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  • Do the distributions and stuff, then split off the pieces that were in the one for $n$ and set them aside for use with the hypothesis and see what happens. – Terra Hyde Sep 05 '15 at 05:03
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    can you use $\LaTeX$ please? – Dr. Sonnhard Graubner Sep 05 '15 at 05:03
  • No idea how to use latex, looking it up. – Jake Burns Sep 05 '15 at 05:28
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    You are stuck because "(n+1) + 3 = (n+1)^2 / 2 + (7/2)(n+1)" is not what the equation means for n or n+1. The equation for $n$ is $$\sum\limits_{k=1}^n (k + 3) = \frac{n^2 + 7n}{2}$$ You get to assume that is true. Now use that to prove the equation for $n+1$: $$\sum\limits_{k=1}^{n+1} (k + 3) = \frac{(n+1)^2 + 7(n+1)}{2}$$ – Paul Sinclair Sep 05 '15 at 05:35
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    By the way, a very useful trick for creating latex expressions: right-click on an expression you would like to copy (or just see how to get some symbol), and select "Show Math As > TEX commands". It gives you everything except for the beginning and ending $s or $$. – Paul Sinclair Sep 05 '15 at 05:40
  • I still get stuck with a second degree polynomial equal to a first degree. To me there appears to be no way to solve it. – Jake Burns Sep 05 '15 at 05:49

1 Answers1

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$\sum\limits_{k=1}^n (k + 3) = \frac{n^2 + 7n}{2}$
For the base case, $n=1$. $$1+3=\frac{8}{2}=4$$
Let us assume that it is true for some $n=p$, then $$\sum\limits_{k=1}^p (k + 3) = \frac{p^2 + 7p}{2}$$ Now it has to be true for all $n=p+1$, then $$\sum\limits_{k=1}^{p+1} (k + 3) = \frac{(p+1)^2 + 7(p+1)}{2}$$ $$\sum_{k=1}^p(k+3)+p+4=\frac{(p+1)^2+7p+7}{2}$$ $$p+4=\frac{(p+1)^2+7p+7}{2}-\frac{p^2+7p}{2}$$ $$p+4=\frac{(2p+8}{2}$$ $$p+4=p+4$$ Which is true. Hence proved.

Aditya Agarwal
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