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Show by induction that the following formulas hold.

$\sum_{i=1}^n n2ⁿ = (n-1)2^{n+1} + 2$

What did a similar problem to this but this one is a little different. I think is because this one has a power n and I'm not sure how to start with this one?

Martin Sleziak
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Alex
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  • Use the dollar signs to format your equations. Copy pasting directly from your assignment never works well. – Hubble Apr 06 '14 at 05:40
  • ok sorry bout that – Alex Apr 06 '14 at 05:41
  • What stops you using the method of http://math.stackexchange.com/questions/741646/use-mathematical-induction-to-prove-that-for-all-integers-n-is-greater-than-or-e – lab bhattacharjee Apr 06 '14 at 05:41
  • I'm new to this site. so i don't know much about that – Alex Apr 06 '14 at 05:42
  • The actual problem is Σⁿ_i=1 n2ⁿ = (n-1)2^((n+1)) + 2 . I'm not sure how to use the dollar sign. this is the closest that I can get – Alex Apr 06 '14 at 05:45
  • To learn how to use the "dollar signs," Google the LaTeX typesetting language. There are tons of beginner's tutorials out there. – phaiakia Apr 06 '14 at 05:48
  • You can consult the help section of this website, or you can also click on "Formatting help" when you post your question. If I understand correctly, you want to find $\sum_{i=1}^n n 2^n$ and prove the result by induction. Shouldn't you be summing up $i$-terms? This doesn't make too much sense to me. Also, would it be possible for you to show us what you have done so far? It's your fifth question on induction in less than 4 hours. Remember that these questions are given to you so you can practice. – Hubble Apr 06 '14 at 05:50
  • This question have been asked (with various minor modifications) on this site several times. See, for example, this question and other questions that you can see there among linked questions. – Martin Sleziak Apr 06 '14 at 06:25
  • BTW you probably wanted to ask about $\sum_{i=1}^n i2^i$ and not about $\sum_{i=1}^n n2^n$. The latter is simply sum which contains $n$ equal summands, each of them equal to $n2^n$. So the result would simply be $n^22^n$. – Martin Sleziak Apr 06 '14 at 06:27
  • iHubble mentioned in a comment that you have asked a lot of questions in a short timespan. I wanted to make sure that you are aware of the quotas 50 questions/30 days and 6 questions/24 hours, so that you can plan posting your questions accordingly. (If you try to post more questions, StackExchange software will not allow you to do so.) For more details see meta. – Martin Sleziak Apr 06 '14 at 06:28

2 Answers2

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Starting with the well known identity: $\sum_{i=1}^n ix^{i-1} = \dfrac{d}{dx}\dfrac{x - x^n}{1 - x} = \dfrac{1 - nx^{n-1} + (n-1)x^n}{(1 -x)^2}$. Then substitute $x = 2$ into the equation, and multiply both sides by $2$ again to get the sought identity.

DeepSea
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I'm assuming you mean to show that $\sum_{i=1}^n i2^i = (n-1)2^{n+1}+2$.

For your base case, $n=1$, you just need to plug in 1: $\sum_{i=1}^1 i 2^i = 1 \cdot 2^1 = 2$ and $(1-1)2^{1+1} + 2 = 2.$

Then for your inductive step, assume $\sum_{i=1}^n i 2^i = (n-1)2^{n+1}+2$, and you want to show that $\sum_{i=1}^{n+1} i 2^i = n2^{n+2}+2$. Hint: rewrite $$\sum_{i=1}^{n+1} i 2^i = (n+1)2^{n+1} + \sum_{i=1}^n i 2^i = (n+1)2^{n+1} + (n-1)2^{n+1}+2$$ and use some basic algebra to rearrange and get what you want.

phaiakia
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