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