Hi I'm stuck on a question, and sadly my coursebook is actually missing the pages I need due to a printing error.

The question reads-

Find the sum of the infinite series:

(5/6)^3 - (5/6)^4 + (5/6)^5 - (5/6) ^ 6 ...

Any help would be great.

Hi edmo,

Another way is......

\(\displaystyle \left(\frac{5}{6}\right)^3-\left(\frac{5}{6}\right)^4+\left(\frac{5}{6}\right)^5-......\)

\(\displaystyle =\left(\frac{5}{6}\right)^3\left(1-\frac{5}{6}+\left(\frac{5}{6}\right)^2-\left(\frac{5}{6}\right)^3+.....\right)\)

\(\displaystyle =\frac{125}{216}\left(1+\left(\frac{5}{6}\right)^2+\left(\frac{5}{6}\right)^4+.......\right)-\frac{125}{216}\left(\frac{5}{6}+\left(\frac{5}{6}\right)^3+\left(\frac{5}{6}\right)^5+.....\right)\)

These are 2 infinite geometric series in brackets.

\(\displaystyle a=first\ term,\ r=geometric\ ratio,\ S_{\infty}=\frac{a}{1-r}\)

For the first one....\(\displaystyle \frac{a}{1-r}=\frac{1}{1-\frac{25}{36}}=\frac{1}{\frac{36-25}{36}}=\frac{36}{11}\)

For the 2nd one.....\(\displaystyle \frac{a}{1-r}=\frac{\frac{5}{6}}{1-\frac{25}{36}}=\frac{5}{6}\ \frac{36}{11}=\frac{30}{11}\)

Hence, the sum of all the terms is

\(\displaystyle \frac{125}{216}\left(\frac{36}{11}-\frac{30}{11}\right)=\frac{125}{216}\ \frac{6}{11}=\frac{125}{11}\ \frac{6}{216}=\frac{125}{11}\ \frac{1}{36}\)