The closed form sum of $$12 \left[ 1^2 \cdot 2 + - id12122214044708, Happypuppy1918

Ответ:

RegencySlayer5304

04.09.2021

Mathematics

Perhaps you know that

$S_2 = \displaystyle\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}6$

and

$S_3 = \displaystyle\sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}4$

Then the problem is trivial, since

$\displaystyle\sum_{k=1}^n k^2(k+1) = S_2 + S_3 \\\\ = \frac{2n(n+1)(2n+1)+3n^2(n+1)^2}{12} \\\\ = \frac{n(n+1)\big((2(2n+1)+3n(n+1)\big)}{12} \\\\ = \frac{n(n+1)\big(4n+2+3n^2+3n\big)}{12} \\\\ = \frac{n(n+1)(3n^2+7n+2)}{12} \\\\ = \frac{n(n+1)(3n+1)(n+2)}{12}$

Then

$12\bigg(1^2\cdot2+2^2\cdot3+3^2\cdot4+\cdots+n^2(n+1)\bigg) = n(n+1)(n+2)(3n+1)$

so that a = 3 and b = 1.

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Ответ:

alicc0529

05.12.2020

Mathematics

148

Step-by-step explanation:

it goes up to down

BCD

ABD

CBE

ADE

AEB

DEA

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The closed form sum of $$12 \left[ 1^2 \cdot 2 + 2^2 \cdot 3 + \ldots + n^2 (n+1) \right]$$
for $n \geq 1$ is $n(n+1)(n+2)(an+b).$ Find $an + b.$

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The closed form sum of $$12 \left[ 1^2 \cdot 2 + 2^2 \cdot 3 + \ldots + n^2 (n+1) \right]$$ for $n \geq 1$ is $n(n+1)(n+2)(an+b).$ Find $an + b.$

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The closed form sum of $$12 \left[ 1^2 \cdot 2 + 2^2 \cdot 3 + \ldots + n^2 (n+1) \right]$$
for $n \geq 1$ is $n(n+1)(n+2)(an+b).$ Find $an + b.$