Question

Evaluate the summation of 20 times 0.5 to the n minus 1 power, from n equals 3 to 12..

Answer 1

You can make use of the formula:
\[ \Large \sum_{n=0}^{t}ar^n=a\frac{1-r^{t+1}}{1-r}\]
And notice that:
\[\Large \sum_{n=3}^{12}0.5^{n-1}=\sum_{n=0}^{12}0.5^{n-1}-\sum_{n=0}^{2}0.5^{n-1}\] This is because the first sum is adding the numbers from 0, 1, 2, 3, 4, ..., 12
the second sum is adding the numbers from 0, 1, 2
So subtracting off the two leaves you only summing from 3, 4, ..., 12.

Thus:\[\Large \begin{align} \sum_{n=3}^{12}20(0.5^{n-1})&=20\sum_{n=3}^{12}0.5^{n-1} \\ &= 20\sum_{n=3}^{12}0.5^n(0.5^{-1}) \\
&=20(0.5^{-1})\sum_{n=3}^{12}0.5^n\\&=40 \left(\sum_{n=0}^{12}0.5^{n}-\sum_{n=0}^{2}0.5^{n} \right) \\
&=40 \left(\frac{1-0.5^{12+1}}{1-0.5}-\frac{1-0.5^{2+1}}{1-0.5} \right)\end{align}\]