Series question please

[Guest]

The question reads:

Find the minimum number of terms which will make the sum of the series 0.125+0.25+0.5+ ... exceed quarter of a million.

Please will you help me to start with this?

Thank you

 

[soroban]

Hello, Amber111!

Find the minimum number of terms which will make
the sum of the series $\displaystyle \frac{1}{8}\,+\,\frac{1}{4}\,+\,\frac{1}{2}\;+\;\cdots$ exceed quarter of a million.

We have a geometric series with first term $\displaystyle a_1\,=\,\frac{1}{8}$ and common ratio $\displaystyle r\,=\,2$

The sum of the first n terms of a geometric series is given by:
\(\displaystyle \L\;\;\;S_n\;=\;a_1\cdot\frac{r^n\,-\,1}{r\,-\,1}\)

Your series has: \(\displaystyle \L\,S_n\;=\;\frac{1}{8}\cdot\frac{2^n\,-\,1}{2\,-\,1}\)

Hence, the sum is: \(\displaystyle \L\,\frac{2^n\,-\,1}{8}\) and it is to be greater than 250,000.


We have: \(\displaystyle \L\,\frac{2^n\,-\,1}{8}\:>\:250,000\;\;\Rightarrow\;\;2^n\:>\:2,000,001\)

Take logs: \(\displaystyle \,\ln(2^n) \:>\:\ln(2,000,001)\;\;\Rightarrow\;\;n\cdot\ln(2)\:>
\:\ln(2,000,001)\)

Hence: \(\displaystyle \L\,n\:>\:\frac{\ln(2,000,001)}{\ln(2)} \:\approx\;20.93\)

Therefore, we need at least 21 terms for the sum to exceed 250,000.

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Check

If $\displaystyle n\,=\,20:\:S_{_{20}}\:=\:\frac{2^{^{20}}\,-\,1}{8}\:=\:131,071.875$

If $\displaystyle n\,=\,21:\:S_{_{21}}\:=\:\frac{2^{^{21}}\,-\,1}{8}\:=\:262,143.875$