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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?

Josefhu15
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    I guess that your professsor, apart from giving you the first step must have also given you some formula for the summation of geometric series. Have you studied it before asking? – KonKan Sep 15 '16 at 04:55
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    Hint: when $a\neq1$, $\sum_{k=0}^na^k=1+a+a^2+\cdots+a^n=\dfrac{a^{n+1}-1}{a-1}$. – Jean-Claude Arbaut Sep 15 '16 at 04:57
  • Besides the professor's hint, another approach would be to factor $3^5$ out of the sum (and then apply the geometric series formula you were presumably given). – carmichael561 Sep 15 '16 at 04:58
  • the summation formulas that he gave to us does not cover anything to the power of n or anything similar to what I have posted. – Josefhu15 Sep 15 '16 at 04:59

2 Answers2

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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 $r\ne 1$).

user2825632
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  • so r = 3 and n = 100 correct? – Josefhu15 Sep 15 '16 at 05:09
  • Yep - for your first sum (from i=0 to i=100). In your second sum (i=0 to i=4), n=4. It should be noted that since the formula includes r^0, and your summation doesn't (you start with 3+9+... instead of 1+3+9+...), you'd also have to subtract 1 from the formula. – user2825632 Sep 15 '16 at 05:12
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$$\sum_{i=1}^{100}3^n=\sum_{i=1}^4 3^n+\sum_{i=5}^{100} 3^n$$ $$3\frac{1-3^{100}}{1-3}=3+3^2+3^3+3^4 +\sum_{i=5}^{100} 3^n$$ $$\frac{3^{101}-3}{2}-120=\sum_{i=5}^{100} 3^n$$

Adi Dani
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