Geometric Progression and Summation

**Punch** · June 9th 2011, 07:34 AM

The sum of the first $n$ terms of a series is given by the expression $(1-3^{-2n})$ .

Find the least value of $k$ such that the sum of the terms from the $k^{th}$ term is less than $\frac{1}{3000}$

$1-3^{2k} < \frac{1}{3000}$

$-3^{-2k} < -\frac{2999}{3000}$

$3^{-2k} > \frac{2999}{3000}$

$-2k < \frac{lg\frac{2999}{3000}}{lg3}$

$k > 0.0001517$

but ans is 5

**Prove It** · June 9th 2011, 08:13 AM

It doesn't say to sum the terms up to the $\displaystyle k^{\textrm{th}}$ term, it says FROM the $\displaystyle k^{\textrm{th}}$ term, presumably up to the $\displaystyle n^{\textrm{th}}$ term.

**Punch** · June 9th 2011, 08:19 AM

Originally Posted by Prove It

It doesn't say to sum the terms up to the $\displaystyle k^{\textrm{th}}$ term, it says FROM the $\displaystyle k^{\textrm{th}}$ term, presumably up to the $\displaystyle n^{\textrm{th}}$ term.

thanks for pointing that out! i'll attempt the question again with this in mind.

**Punch** · June 9th 2011, 08:36 AM

$\sum_{n = k}^N(1-3^{-2n}) < \frac{1}{3000}$

$\sum_{n = k}^N1-\sum_{n = k}^N(3^{-2n}) < \frac{1}{3000}$

$(N-k+1)(1)-(3^{-2k}+3^{-2(k+1)}+...+3^{-2N}) < \frac{1}{3000}$

$N-k+1-\frac{3^{-2k}(1-9^{N-k+1})}{1-9}<\frac{1}{3000}$

but N doesnt seem to be cancelling out

**Punch** · June 10th 2011, 06:44 AM

Originally Posted by Punch

$\sum_{n = k}^N(1-3^{-2n}) < \frac{1}{3000}$

$\sum_{n = k}^N1-\sum_{n = k}^N(3^{-2n}) < \frac{1}{3000}$

$(N-k+1)(1)-(3^{-2k}+3^{-2(k+1)}+...+3^{-2N}) < \frac{1}{3000}$

$N-k+1-\frac{3^{-2k}(1-9^{N-k+1})}{1-9}<\frac{1}{3000}$

but N doesnt seem to be cancelling out

am i on the right track?

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