# Thread: numerical value of summation

1. ## numerical value of summation

need your guide on this expression.

thanks

2. ## Re: numerical value of summation

Originally Posted by rcs
need your guide on this expression.

thanks
\displaystyle \displaystyle \begin{align*} \sum_{k = 10}^{200}{\frac{1}{k^2 + 3k}} &= \sum_{k = 10}^{200}{\frac{1}{k(k + 3)}} \end{align*}

Applying Partial Fractions...

\displaystyle \displaystyle \begin{align*} \frac{A}{k} + \frac{B}{k + 3} &\equiv \frac{1}{k(k + 3)} \\ \frac{A(k + 3) + B\,k}{k(k + 3)} &\equiv \frac{1}{k(k + 3)} \\ A(k + 3) +B\,k &\equiv 1 \\ A\,k + 3A + B\,k &\equiv 1 \\ (A + B)k + 3A &\equiv 0k + 1 \\ 3A = 1 \textrm{ and } A + B &= 0 \\ A = \frac{1}{3} \textrm{ and } B &= -\frac{1}{3} \end{align*}

So

\displaystyle \displaystyle \begin{align*} \sum_{k = 10}^{200}{\frac{1}{k(k + 3)}} &= \sum_{k = 10}^{200}\left[ \frac{1}{3k} - \frac{1}{3(k + 3)} \right] \\ &= \frac{1}{3}\sum_{k = 10}^{200}\left(\frac{1}{k}\right) - \frac{1}{3} \sum_{k = 10}^{200}\left(\frac{1}{k + 3}\right) \\ &= \frac{1}{3}\left(\frac{1}{10} + \frac{1}{11} + \frac{1}{12} + \frac{1}{13} + \frac{1}{14} + \dots + \frac{1}{200}\right) - \frac{1}{3}\left(\frac{1}{13} + \frac{1}{14} + \dots + \frac{1}{200} + \frac{1}{201} + \frac{1}{202} + \frac{1}{203}\right) \\ &= \frac{1}{3}\left(\frac{1}{10} + \frac{1}{11} + \frac{1}{12} - \frac{1}{201} - \frac{1}{202} - \frac{1}{203}\right) \end{align*}

You can simplify this further if you like...

3. ## Re: numerical value of summation

Thank you BEST MHF Helper Sir. ProveIt ... King of Proofs