Proof by induction, summation series
- Thread starter ka923
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pka
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@ka923, having nine other posts you know the routine here.
You show what you have done. You ask questions about your work. WE try to help you.
Steven G
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The answer to your question really depends on what Z0+ means in your text/class. You can always check to see if both sides are equal when n=0 and if it is, then it really doesn't matter if you start with n=0or n=1
Normally Z+ means positive integers but that subscript of 0 throws me off a bit
Steven G
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The 2nd to the last line is not true. The 2nd summation (if you insist on using sigma notion) should have r going from k+1 to k+1 as you are adding on the last term. You are adding the 1st 2 terms twice. Why??
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Steven G
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But you added the 1st two terms! Also, as I hope you already noticed, the 1st term and the last term may not be equal!
Steven G
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Did you mean---i tried adding my result from r=k to (k+1)??
If yes, why are you adding the results for r=k in both summations? Just add on the results for r=k+1 to the summation from r = 0to r =k (or from r=1to r=k)
Steven G
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The last term is simply (k+1)^4
pka
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Suppose that \(\displaystyle \sum\limits_{k = 1}^N {{k^4}} = \frac{1}{{30}}N(N + 1)(2N + 1)(3{N^2} + 3N - 1)\) IS TRUE.
Then
\(\displaystyle \begin{align*}\sum\limits_{k = 1}^{N + 1} {{k^4}} &= \sum\limits_{k = 1}^N {{k^4}} + {(N + 1)^4} \\&=\frac{1}{{30}}N(N + 1)(2N + 1)(3{N^2} + 3N - 1)+(N+1)^4 \end{align*}\)
What can you do with that?