+0

# anyone????

0
145
3

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

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"

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

Jun 1, 2020
#2
0

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
0

you don't find n; the final answer is 3n+1

Guest Jun 8, 2020