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(x±a)(x±b) = x²±(a+b)x+ab.

evaluate summ...

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evaluate summation of 1/i^(4m+k) K=1 to 2015 m belongs to N

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\sum_{r = 1}^{n} [\sum_{k = 1}^{r} k] {[{log}_{1 / 2} \sqrt{(4 x - x^{2})}]}_{1 / 2}^{r}

x

for which summation is a finite number as

n \to \infty

Q. If k is and n be + ive integers and

s_{k} = 1^{k} + 2^{k} + 3^{k} . . . + n^{k},

\sum_{r = 1}^{m}^{m + 1} C_{r} S_{r} =, (n + 1)^{m + 1} - (n + 1) .

s_{4} .

^{n - 1} C_{r} = {(K^{2} - 3)}^{n} C_{r + 1^{'}}

k

^{n - 1} C_{r} = (k^{2} - 3)^{n} C_{r + 1},

then k belongs to the interval

^{n} C_{r - 1} = (k^{2} - 8) (^{n + 1} C_{r})

k

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(x±a)(x±b) = x²±(a+b)x+ab.

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