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I'm suppose to find the value of

$$ \sum\limits_{i=5}^{100}(3)^n $$

My professor gave me the first step to this which is $$ \sum\limits_{i=1}^{100}(3)^n - \sum\limits_{i=1}^{4}(3)^n $$

and I honestly can't figure out what to do next. I did not find this question posted anywhere else so if it is sorry for posting and can you redirect me?

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2 Answers

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$$\sum_{i=1}^{100}3^n=\sum_{i=1}^4 3^n+\sum_{i=5}^{100} 3^n$$ $$3\frac{1-3^{100}}{1-3}=3+3^2+3^3+3^4 +\sum_{i=5}^{100} 3^n$$ $$\frac{3^{101}-3}{2}-120=\sum_{i=5}^{100} 3^n$$

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Here's a hint. Let $1+r+r^2+...+r^n = x$ and we want to find $x$. Then we see:

$$1+r+r^2+...+r^n = x$$ $$(1+r+r^2+...+r^n)(1-r) = x(1-r)$$ $$(1+r+r^2+...+r^n)-(r+r^2+...+r^n+r^{n+1})=x(1-r)$$ $$1-r^{n+1} = x(1-r)$$ $$\frac{1-r^{n+1}}{1-r} = x$$ $$1+r+r^2+...+r^n=\frac{1-r^{n+1}}{1-r}$$

Can you figure out what $r$ and $n$ are in your problem? (Note that the formula above only works for $r\ne 1$).

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