Answered

IDNLearn.com connects you with a community of experts ready to answer your questions. Find in-depth and accurate answers to all your questions from our knowledgeable and dedicated community members.

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.

We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for visiting IDNLearn.com. We’re here to provide accurate and reliable answers, so visit us again soon.