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How many terms of the geometric series 7, 7/2, 7/4, 7/8, ... must be taken in order that the sum may differ from the sum to infinity by less than 0.01? (Ans: 11 terms)

Some help, please?

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What have you tried? –  jspecter Jun 28 '11 at 13:16
    
@jspecter: I don't even know how to start doing it when I saw this question. –  Sophia Jun 28 '11 at 13:28

3 Answers 3

Do you know how to sum a geometric series? Basically you are being asked how late to start so that the sum from there on is less than 0.01. So if you write the sum starting from $n$ to $\infty$ you should be able to find $n$ that meets this. A geometric series stays geometric if you cut off some of the first terms.

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HINT:

We know how to sum geometric series ($\dfrac{a_0}{1-r}$ where $a_0$ is the first term and $r$ is the ratio). We also know how to sum partial geometric series, by either directly adding them up (applicable here) or by recalling some useful formulas.

You could subtract these, or you could realize that the difference is just another geometric series with a different starting point.

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Hint: Let $x=1/2$ in the formula $$ 1 + x +x^2 + \cdots + x^n = \frac{{1 - x^{n + 1} }}{{1 - x}}. $$

EDIT:

Let $S_n$ be the $n$th partial sum, so that $$ S_1 = 7, $$ $$ S_2 = 7 + 7/2 = 7(1+1/2), $$ $$ S_3 = 7 + 7/2 + 7/4 = 7(1+1/2+1/4), $$ from which you can conclude that $$ S_n = 7 (1+1/2+\cdots+1/2^{n-1}). $$ Further, let $S$ denote the sum to infinity, so that $$ S = 7 (1+1/2+1/4+\cdots) = 7 \cdot 2 = 14. $$ You are asked to find the least $n$ such that $$ S - S_n < 0.01. $$ Noting that $$ S_n = 7 (1+1/2+\cdots+1/2^{n-1}) = 7 \frac{{1 - (1/2)^n }}{{1 - 1/2}} = 14(1 - 1/2^n ), $$ it remains to find the least $n$ such that $$ 14 - 14(1 - 1/2^n ) < 0.01, $$ or $$ 14/2^n < 0.01, $$ or $$ 2^n > 1400. $$ Since $2^{10} = 1024 < 1400$ and $2^{11} = 2048 > 1400$, the answer is $11$.

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Furthermore, it doesn't ask for the least $n$, just some $n$ that works. In this case we can find the least $n$ as seen, but in many cases it is more convenient to work with estimates, rather than exact formulas for the partial sums, and then we can still find an $n$ that works. –  GEdgar Jun 28 '11 at 14:33
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@GEdgar: "the least $n$" (rather than "some $n$") corresponds to "must be taken" (rather than "can be taken"). –  Shai Covo Jun 28 '11 at 14:53
    
And anyway, the OP mentioned that the answer is "$11$ terms". –  Shai Covo Jun 28 '11 at 14:56

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