Sequence and series

Feb 2018
60
3
Iran
20191205_182411.jpg
How can I solve this problem easily? please explain step by step
 

v8archie

Math Team
Dec 2013
7,709
2,677
Colombia
  1. Decompose the fraction by the method of partial fractions.
  2. Spot the telescoping series and thus remove many terms of the sum.
  3. Sum the remaining terms.
 
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Dec 2015
939
122
Earth
Substitute for \(\displaystyle 2k=t \; \) , the sum becomes \(\displaystyle s=\sum_{t=2}^{60} \dfrac{2}{t^2 -1 } \).

Solve the equation for \(\displaystyle p \: \) : \(\displaystyle \dfrac{2}{t^2 -1 }=p[\dfrac{1}{t-1}-\dfrac{1}{t+1} ] \; \Rightarrow \) \(\displaystyle p=1\).

Now the expression inside the sum(general term) becomes \(\displaystyle \dfrac{1}{t-1}-\dfrac{1}{t+1}\).

\(\displaystyle s =\displaystyle \sum_{t=2}^{60} [\dfrac{1}{t-1}-\dfrac{1}{t+1}]=\sum_{t=2}^{60}\dfrac{1}{t-1} -\sum_{t=2}^{60}\dfrac{1}{t+1}\).

\(\displaystyle s=[1+1/2+1/3+\dotsc +1/59]-[1/3+1/4+1/5+\dotsc +1/59 +1/60+1/61]=1+1/2-1/60-1/61\).
 

skeeter

Math Team
Jul 2011
3,212
1,734
Texas
If $2k=t$, then $t$ cannot be an odd value ...

\(\displaystyle \sum_{k=1}^{30} \dfrac{1}{2k-1} - \dfrac{1}{2k+1} = \dfrac{60}{61}\)
 
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Dec 2015
939
122
Earth
Correct version : \(\displaystyle \sum_{t=2k}^{60} 1/(t-1) - 1/(t+1)=[1+1/3+1/5+...+1/59]-[1/3+1/5+...+1/61]=1-1/61=60/61\).