The question is asking to find the sum of a geometric series up to term n, where the first value of r is 1. The expression is $$ \sum_{r=1}^n \ 4r^3-3r^2+r$$. I am aware of the general formula for the sum of a geometric series, but don't know what to use as the common ratio - is there any way to figure out the common ratio from the equation?
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2What are you summing? – Joshua Wang Jan 28 '21 at 02:13
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WA answer about $\sum\limits_{r=1}^n 4r^3-3r^2+r$ @oscar6721 – Alexey Burdin Jan 28 '21 at 02:16
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@AlexeyBurdin how would you go about showing that this is the case, apart from just putting values into the expression? – oscar6721 Jan 28 '21 at 02:29
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Let $f(r)=4r^3-3r^2+r$ and $F(n)=An^4+Bn^3+Cn^2+Dn+E$ and $F(n)-F(n-1)\equiv f(n)$, then equate the latter term-wise, viola. – Alexey Burdin Jan 28 '21 at 02:32
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This is NOT a geometric series by the way – Joshua Wang Jan 28 '21 at 02:47
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First, we use the identities:
$$\begin{align}&\sum_{i = 1}^{n}i^{3} = \bigg(\frac{n(n +1)}{2}\bigg)^{2} \\&\sum_{i = 1}^{n}i^{2} = \frac{n(n + 1)(2n + 1)}{6} \\&\sum_{i = 1}^{n}i = \frac{n(n + 1)}{2}\end{align}$$
These can be easily proved by induction. Then, for your sum we have:
$$\begin{align}\sum_{r=1}^{n}4r^{3}-3r^{2} + r&= 4\sum_{r = 1}^{n}r^{3}-3\sum_{r=1}^{n}r^{2}+\sum_{r=1}^{n}r \\&= n^{2}(n+1)^{2}-\frac{n(n+1)(2n+1)}{2}+\frac{n(n + 1)}{2} \\&=n^{2}(n + 1)^{2}-\frac{n(n + 1)(2n + 1 - 1)}{2} \\&= n^{2}(n + 1)^{2} - n^{2}(n + 1) \\&= \boxed{n^{4} + n^{3}}\end{align}$$
19aksh
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Joshua Wang
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If you are interested in how these identities are derived, see https://math.stackexchange.com/questions/155166/finite-sum-of-power. – Joshua Wang Jan 28 '21 at 02:46