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I was asked to show by induction that " $\forall n \in \mathbb{N},-1 \cdot3+2\cdot 5-3\cdot7+...+2n(4n+1)= \alpha n $" and find the value of the constant $\alpha$.

First, I rewrote the sum on the left side as $$ \sum_{k=0}^{2n} (-1)^{k}\cdot k(2k+1)$$

Next, I tried to check the cases $n=1$ and $n=2$ in order to find the constant, but here is where my problem arises. For $n=1$ the value of the sum is 7, for $n=2$ is 22 and for $n=3$ the result is 45. Due to the above, it's impossible that this follows a linear partial sum for 2n. Am I correct or am I missing something?

Thanks for reading.

George25
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    In your sum, the second factor $k+1$ should be $2k+1$. – Andreas Blass Apr 14 '20 at 00:23
  • Welcome to Mathematics Stack Exchange. How about $\alpha=4n+3$? – J. W. Tanner Apr 14 '20 at 00:25
  • In any case, you are correct that a sum like yours will not result simplify into $\alpha n$ for some scalar $\alpha$. It will happen to be quadratic. You might consider splitting the sum into two to prove this: $2\sum\limits_{k=0}^{2n}(-1)^k\cdot k^2 + \sum\limits_{k=0}^{2n}(-1)^kk$. If you look around you can find posts like this one for each of those individual sums and them simplify. – JMoravitz Apr 14 '20 at 00:28
  • @J.W.Tanner The problem statement says "constant $\alpha$" – JMoravitz Apr 14 '20 at 00:31
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    @JMoravitz The part about "constant" was not in the original version of the question but was edited in later. – Andreas Blass Apr 14 '20 at 00:34
  • @AndreasBlass That's correct. Sorry about that. – George25 Apr 14 '20 at 00:35
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    Alternatively, you can group each negative term with the immediately following positive term. That simplifies things down to an easy summation. – Andreas Blass Apr 14 '20 at 00:35
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    Thank you all, I think I'll just send an email to the teacher's assistant about this. – George25 Apr 14 '20 at 00:37

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