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Chain Rule of Differentiation

If ∑r=1n rr...

Question

If $n \sum r = 1 r (r + 1) (2 r + 3) = a n + b n^{3} + c n^{2} + d n + e$ , then

a = 1 / 2

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b = 8 / 3

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c = 9 / 2

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e = 0

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Solution

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\sum_{r = 1}^{n} (r) (r + 1) (2 r + 3) = a n^{4} + b n^{3} + c n^{2} + d n + e

n \sum k = 1 k \sum m = 1 m^{2} = a n^{4} + b n^{3} + c n^{2} + d n + e, then

\sum_{k = 1}^{n} (\sum_{m = 1}^{k} m^{2}) = a n^{4} + b n^{3} + c n^{2} + d n + e

If $\sum_{k = 1}^{n}$ $\sum_{m = 1}^{k}$ $m^{2}$ = a $n^{4}$ + b $n^{3}$ + c $n^{2}$ + dn + e. then b =

S_{n} = (2 + 2) + (2^{2} + 5) + (2^{3} + 10) + (2^{4} + 17) + . . . . . . . .

n

S_{n} = 2^{n + A} + B n^{3} + C n^{2} + D n + E

n \in N

A, B, C, D, E

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