1. Geometric Progression

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

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

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

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

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

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

$\displaystyle k > 0.0001517$

but ans is 5

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

3. Originally Posted by Prove It
It doesn't say to sum the terms up to the $\displaystyle \displaystyle k^{\textrm{th}}$ term, it says FROM the $\displaystyle \displaystyle k^{\textrm{th}}$ term, presumably up to the $\displaystyle \displaystyle n^{\textrm{th}}$ term.
thanks for pointing that out! i'll attempt the question again with this in mind.

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

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

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

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

but N doesnt seem to be cancelling out

5. Originally Posted by Punch
$\displaystyle \sum_{n = k}^N(1-3^{-2n}) < \frac{1}{3000}$

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

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

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

but N doesnt seem to be cancelling out
am i on the right track?