# Mathematical Induction

• Aug 31st 2010, 01:40 PM
zirly
Mathematical Induction
Hi guys, im new to the forum and was wandering if you guys could help me with a question on mathematical induction, its part of an assignment I have to complete and its been bugging me for a while, thanks in advance if you could help me out!

The question is an image i will upload.
Attachment 18790

May i add i have tried and it has worked for the base case.
• Aug 31st 2010, 01:50 PM
Plato
Write as $\displaystyle \displaystyle\frac{1} {{\left( {3i - 1} \right)\left( {3i + 2} \right)}} = \frac{1} {{3\left( {3i - 1} \right)}} - \frac{1} {{3\left( {3i + 2} \right)}}$.
• Aug 31st 2010, 03:13 PM
zirly
sorry for asking but what difference does it make if the lhs is changed? i thought the main focus was the rhs
• Aug 31st 2010, 03:31 PM
Plato
Quote:

Originally Posted by zirly
sorry for asking but what difference does it make if the lhs is changed? i thought the main focus was the rhs

Well for one it turns the sums into collapsing sums.
$\displaystyle \displaystyle \sum\limits_{i = 1}^3 {\frac{1} {{\left( {3i - 1} \right)\left( {3i + 2} \right)}}} = \sum\limits_{}^{} {\left[ {\frac{1} {{3\left( {3i - 1} \right)}} - \frac{1} {{3\left( {3i + 2} \right)}}} \right]}$
$\displaystyle \displaystyle = \left( {\frac{1} {6} - \frac{1} {{15}}} \right) + \left( {\frac{1} {{15}} - \frac{1} {{24}}} \right) + \left( {\frac{1} {{24}} - \frac{1} {{33}}} \right) = \left( {\frac{1} {6} - \frac{1} {{33}}} \right)$