If the sum of the series ((1)/(2)-(1)/(3))+(12^2-(1)/(2 3)+13^2)+(12^3-12^2 3+12 3^2-13^3)…

Question

If the sum of the series

$(\frac{1}{2}-\frac{1}{3})+(\frac{1}{{2}^{2}}-\frac{1}{2\cdot 3}+\frac{1}{{3}^{2}})+(\frac{1}{{2}^{3}}-\frac{1}{{2}^{2}\cdot 3}+\frac{1}{2\cdot {3}^{2}}-\frac{1}{{3}^{3}})+$

$(\frac{1}{{2}^{4}}-\frac{1}{{2}^{3}\cdot 3}+\frac{1}{{2}^{2}\cdot {3}^{2}}-\frac{1}{2\cdot {3}^{3}}+\frac{1}{{3}^{4}})+\ldots$

is $\frac{\alpha }{\beta }$, where $\alpha $ and $\beta $ are co-prime, then $\alpha +3\beta $ is equal to \_\_\_\_\_\_\_\_\_\_.

Answer: 7

Accepted Answer

We can rewrite the given series as follows :

$S=(\frac{1}{2}-\frac{1}{3})+(\frac{1}{4}-\frac{1}{6}+\frac{1}{9})+(\frac{1}{8}-\frac{1}{12}+\frac{1}{18}-\frac{1}{27})+(\frac{1}{16}-\frac{1}{24}+\frac{1}{36}-\frac{1}{54}+\frac{1}{81})+\ldots$

The first few terms of the series are :

$S=(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots )-(\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\ldots )+(\frac{1}{9}+\frac{1}{18}+\frac{1}{36}+\ldots )-(\frac{1}{27}+\frac{1}{54}+\ldots )+\ldots$

We can now see that each group of terms forms a geometric series with a common ratio of $\frac{1}{2}$:

$S=\frac{\frac{1}{2}}{1-\frac{1}{2}}-\frac{\frac{1}{3}}{1-\frac{1}{2}}+\frac{\frac{1}{9}}{1-\frac{1}{2}}-\frac{\frac{1}{27}}{1-\frac{1}{2}}+\ldots$

The series can be rewritten as :

$S=2(\frac{1}{2}-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+\ldots )$

Now, we can simplify and rewrite the series inside the parentheses as:

$S=2[\frac{1}{2}+(-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}+\ldots )]$

The series inside the parentheses is an infinite geometric series with the first term $a=-\frac{1}{3}$ and the common ratio $r=-\frac{1}{3}$:

$S=2(\frac{1}{2}+\frac{a}{1-r})=2(\frac{1}{2}+\frac{-\frac{1}{3}}{1+\frac{1}{3}})=2(\frac{1}{2}+\frac{-\frac{1}{3}}{\frac{4}{3}})$

= $2(\frac{1}{2}-\frac{1}{3}	imes \frac{3}{4})$

$=2(\frac{1}{2}-\frac{1}{4})=2(\frac{1}{4})=\frac{1}{2}$

Thus, the sum of the series is $\frac{1}{2}$, and $\alpha =1$ and $\beta =2$ are co-prime.

Therefore, $\alpha +3\beta =1+6=7$

Solutions · 1