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