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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 $an + b.$

The Closed Form Sum Of 12 Left 12 Cdot 2 22 Cdot 3 Ldots N2 N1 Right For N Geq 1 Is Nn1n2anb Find An B class=

Sagot :

LammettHashidn

Perhaps you know that

[tex]S_2 = \displaystyle\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}6[/tex]

and

[tex]S_3 = \displaystyle\sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}4[/tex]

Then the problem is trivial, since

[tex]\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}[/tex]

Then

[tex]12\bigg(1^2\cdot2+2^2\cdot3+3^2\cdot4+\cdots+n^2(n+1)\bigg) = n(n+1)(n+2)(3n+1)[/tex]

so that a = 3 and b = 1.

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