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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

 Jun 1, 2020
 #1
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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"

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

.
 Jun 1, 2020
 #2
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hi other guest!

first of all, thank you for your given solution (however it was incorrect.) i did the same thing as you before with a and b and they were 3 and 1, but how do i find n?

Guest Jun 1, 2020
 #3
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you don't find n; the final answer is 3n+1

Guest Jun 8, 2020

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