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Show by induction that for all integers n $\ge$ 1 $$ \sum_{i=1}^n i3^i = \frac{3(2n3^n-3^n+1)}{4} $$

Starting with n = 1 will give me LHS = 3 and RHS = 3.

Inserting n = p gives $$\sum\limits_{i=1}^p i3^i = \frac{3(2p3^p-3^p+1)}{4} $$

Going to n = p+1 gives me LHS $$\frac{3(2p3^p-3^p+1)}{4} +(p+1)3^{p+1}$$

I tried to expand the RHS for $$\frac{3(2(p+1)3^{p+1}-3^{p+1}+1)}{4}$$ so it becomes $$\frac{3(2p+2)3^{p+1}-3^{p+1}+1)}{4}$$ then $$\frac{3(p(2+2)3^{p+1}-3^{p+1}+1)}{4}$$ until $$\frac{3(p(2+2)3*3^{p}-3*3^{p}+1)}{4}$$

Is this the right way to expand this formula?

addde
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1 Answers1

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There seem to be numerous typos, either in terms of missing parentheses, or mismatched variables in your solution.

In particular, the claim is written as $\sum\limits_{i=1}^n i3^i=\frac{3(2i3^i-3i+1)}{4}$. Surely, you mean to have $n$'s on the RHS instead of $i$'s.

The line that begins "inserting $n=p$ gives", shouldn't that be $\sum\limits_{i=1}^p i3^i = \frac{3(2p3^p-3p+1)}{4}$. Why do you write $p3^p$?

After the phrase "so it becomes" you have $\frac{3(2p+2)3^{p+1}-3^{p+1}+1)}{4}$ which has a total of one left parenthesis and two right parenthesis. Surely, you meant to say $\frac{3\color{red}{(}(2p+2)3^{p+1}-3^{p+1}+1)}{4}$

The next line which begins "then" you have $\frac{3(p(2+2)3^{p+1}-3^{p+1}+1)}{4}$. How did that step go? $(2p+2)\neq p(2+2)$ in general (only for the one case when $p=1$)

You have made a good start, but there is still more manipulation to do before you finally get the LHS to appear equal the RHS. It is usually easier to put things into a fraction than take things out of a fraction, so I would suggest continuing to work on the LHS a bit more.

For information on how to format an induction proof, I recommend viewing this answer.

For the beginnings of a better worded proof:

Claim: $\sum\limits_{i=1}^ni3^i = \frac{3(2n3^n-3n+1)}{4}$ for all $n\geq 1$

Proof: The claim holds for $n=1$.

Let us assume for our induction hypothesis that $\sum\limits_{i=1}^pi3^i = \frac{3(2p3^p-3p+1)}{4}$ for some $p\geq 1$. Then for $p+1$ we have:

$$\begin{array}{lll} \sum\limits_{i=1}^{p+1}i3^i &= (p+1)3^{p+1}+\sum\limits_{i=1}^p i3^i\\ &=(p+1)3^{p+1}+\frac{3(2p3^p-3p+1)}{4}&\text{by induction hypothesis}\\ &=\frac{4p3^{p+1}+4\cdot 3^{p+1}}{4}+\frac{3(2p3^p-3p+1)}{4}\\ &=\frac{4p3^{p+1}+4\cdot 3^{p+1} + 2p3^{p+1}-9p+3}{4}\\ &=\frac{6p3^{p+1}+4\cdot 3^{p+1} - 9p+3}{4}\\ &\vdots \end{array}$$

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JMoravitz
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  • Yes i have edited so it is N on the RHS and changed insertining n = p as you suggested.

    $$\frac{3(p(2+2)3^{p+1}-3^{p+1}+1)}{4} $$ was a guess from me since i couldnt figure out where to go.

     – addde Aug 12 '15 at 06:57
  • @addde added the beginnings of a proof. Remember that you can always multiply by one and it won't change anything and you can always add zero and it won't change anything. The trick sometimes is what "one" and "zero" you use. (By "adding zero" that could mean adding and subtracting the same thing, like adding $(2p+1-2p-1)$). It also helps sometimes to break pieces apart into multiple smaller pieces, like $4\cdot 3^{p+1}=2\cdot 3^{p+1} + 2\cdot 3^{p+1}$ etc – JMoravitz Aug 12 '15 at 07:12
  • RHS should be $$\sum_{i=1}^n i3^i = \frac{3(2n3^n-3^n+1)}{4}$$ though. – addde Aug 12 '15 at 08:21
  • Which is why I put dots. I didn't finish the proof. There's only one or two steps left to complete it though. – JMoravitz Aug 12 '15 at 16:03