I've posted this question two times, but no one has answered so I'm going to post it one last time.. :(

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.

https://web2.0calc.com/questions/the-closed-form-sum-of-for-is-find-an-b

https://web2.0calc.com/questions/please-help_99797

Guest Jun 1, 2020

#1**0 **

Hello Guest,

I will give you hints to try it then the solution.

-Let n be any number greater than 1 (E.g. 4)

-Substitute in the "Closed-form sum"

-Substitute in the formula given \(n(n+1)(n+2)(an+b)\)

Further hint:

-Set them equal

Solution:

let n=4

\(4(4+1)(4+2)(4a+b)=12(1^2*2+2^2*3+3^2*4+4^2*5)\)

\(4(5)(6)(4a+b)=1560\)

\(480a+120b=1560\)

\(12a+3b=39\)

\(4a+b=13\)

Now, let n=3 and obtain another equation to solve for a and b

\(3(4)(5)(3a+b)=12(1^2*2+2^2*3+3^2*4)\)

\(60(3a+b)=600\)

\(3a+b=10\)

Now we have system of two equations:

\(4a+b=13\)

\(3a+b=10\)

multiply the second equation by (-1) and add it to the first equation.

\(a=3\)

\(b=1\)

so the formula for this sequence:

\(n(n+1)(n+2)(3n+1)\)

We can verify this formula by letting n to be equal to any number, for instance, n=6

\(6(7)(8)(19)=6384\)

Now using the general "Closed form sum"

\(12(1^2*2+2^2*3+3^2*4+4^2*5+5^2*6+6^2*7)=6384\)

Guest Jun 1, 2020