$$\sum_{n=2}^\infty \frac{(4\cdot3^n)}{5^{n-2}}$$
I know the answer is 90, I have an answer key and have plugged the series in to a few (step-less) online calculators. However I'd like to know how to solve this alone.
How would I determine the sum of this infinite series?
0
$\begingroup$
calculus
sequences-and-series
3 Answers
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.(\begin{array}{l}\sum\limits_2^\infty {\frac{{4 \times {3^n}}}{{{5^{n - 2}}}}} = {\sum\limits_2^\infty {\left( {\frac{3}{5}} \right)} ^n} \times 25 \times 4 = 100 \times {\sum\limits_2^\infty {\left( {\frac{3}{5}} \right)} ^n} = 100 \times \left( {{{\sum\limits_0^\infty {\left( {\frac{3}{5}} \right)} }^n} - {{\sum\limits_0^1 {\left( {\frac{3}{5}} \right)} }^n}} \right)\\ = 100 \times \left( {\frac{5}{2} - \frac{8}{5}} \right) = 90\\Note{\rm{ : }}{\sum\limits_2^\infty {\left( {\frac{3}{5}} \right)} ^n} = {\sum\limits_0^\infty {\left( {\frac{3}{5}} \right)} ^n} - {\sum\limits_0^1 {\left( {\frac{3}{5}} \right)} ^n} = \frac{1}{{1 - \frac{3}{5}}} - \left( {1 + \frac{3}{5}} \right) = \frac{5}{2} - \frac{8}{5} = \frac{9}{{10}}\end{array}).
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HINT: write your sum in the form $$100\sum_{n=2}^{\infty}\left(\frac{3}{5}\right)^n$$ and use the geometric sum formula.
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$\require{cancel}\sum_{n=2}^\infty \frac{(4\cdot3^n)}{5^{n-2}}=4\cdot 5^2\sum_{n=2}^\infty \frac{3^n}{5^{n}}=4\cdot 5^2\sum_{n=2}^\infty (\frac{3}{5})^n$
GP where $a=\frac{9}{25}, r=\frac{3}{5}$
Sum of GP=$ a\frac{1-r^n}{1-r}$ , as $n\to \infty, r^n\to 0$
$\therefore$ sum of given GP=$4 \cdot 5^2 \left( a/(1-r)\right)=4 \cdot 5^2 \cdot \frac{9}{25}\cdot \frac{1}{1-3/5}=\cancelto{2}4 \cdot \cancel{5^2} \cdot \frac{9}{\cancel{25}}\cdot \frac{5}{\cancel{2}}=90$