\(~~~\displaystyle ~~\sum_{k=4}^{n} \{~~f(k-2) - f(k)~~\}\)

\(=~~\displaystyle \sum_{k=4}^{n}f(k-2) - \sum_{k=4}^{n}f(k) \)

\(=~~\displaystyle \sum_{k=2}^{n-2}f(k) - \sum_{k=4}^{n}f(k) \)

\(=~~\displaystyle f(2) + f(3) +\sum_{k=4}^{n-2}f(k) - \sum_{k=4}^{n-2}f(k) - f(n-1) - f(n) \)

\(=~~\displaystyle f(2) + f(3) + ............... 0 ............................. - f(n-1) - f(n)\)

\(\displaystyle NOTE:-Change~of~boundareys \) \(\displaystyle \sum_{k=m}^{n}f(k) = \sum_{k=m+p}^{n+p}f(k-p)…p~is~any~intiger.\)

\(\displaystyle \sum_{k=m}^{n}f(k) = \sum_{k=m}^{r}f(k) + \sum_{k=r+1}^{n}f(k)->splitting~the~range~into~~m~to~r~~and~~~(r+1)~to~ n.\)

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

## Comments

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

Cheers – John Muradeli · 3 years ago

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– Niranjan Khanderia · 3 years ago

Thanks. I have now used \( \displaystyle \color\red{Latex} \)Log in to reply