Telescopic Series

     ∑k=4n{  f(k−2)−f(k)  }~~~\displaystyle ~~\sum_{k=4}^{n} \{~~f(k-2) - f(k)~~\}
=  ∑k=4nf(k−2)−∑k=4nf(k)=~~\displaystyle \sum_{k=4}^{n}f(k-2) - \sum_{k=4}^{n}f(k)
=  ∑k=2n−2f(k)−∑k=4nf(k)=~~\displaystyle \sum_{k=2}^{n-2}f(k) - \sum_{k=4}^{n}f(k)
=  f(2)+f(3)+∑k=4n−2f(k)−∑k=4n−2f(k)−f(n−1)−f(n)=~~\displaystyle f(2) + f(3) +\sum_{k=4}^{n-2}f(k) - \sum_{k=4}^{n-2}f(k) - f(n-1) - f(n)
=  f(2)+f(3)+...............0.............................−f(n−1)−f(n)=~~\displaystyle f(2) + f(3) + ............... 0 ............................. - f(n-1) - f(n)

NOTE:−Change of boundareys\displaystyle NOTE:-Change~of~boundareys ∑k=mnf(k)=∑k=m+pn+pf(k−p)…p is any intiger.\displaystyle \sum_{k=m}^{n}f(k) = \sum_{k=m+p}^{n+p}f(k-p)…p~is~any~intiger.
∑k=mnf(k)=∑k=mrf(k)+∑k=r+1nf(k)−>splitting the range into  m to r  and   (r+1) to n.\displaystyle \sum_{k=m}^{n}f(k) = \sum_{k=m}^{r}f(k) + \sum_{k=r+1}^{n}f(k)->splitting~the~range~into~~m~to~r~~and~~~(r+1)~to~ n.

We change the first summation from 4 through n to 2 through n-2 so that the expression inside becomes X^k for both summation. Then separate out summations between k=4 and k=n-2 which cancels out to 0. We are left with remaining first TWO terms from the first summation and last TWO from the second, since there is a difference of 2 (p=2) in the start of the two summations. For other p values this should be adjusted suitably

Note by Niranjan Khanderia
7 years, 2 months ago

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  Easy Math Editor

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Comments

Um... Consider using latex. All this "^" stuff makes it really confusing.

Cheers

John M. - 6 years, 11 months ago

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Thanks. I have now used \displaystyle \color\red{Latex}

Niranjan Khanderia - 6 years, 11 months ago
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