Find value of this sum

[xone306]

Hello, got a question that I'm struggling with and can find no clear explanations on how to do it anywhere online:
Find the sum of:

n
E (i+1)(i+2)
i=1

(where E=sigma)

I figured that you should foil the brackets, making it i²+3i+2, then break that up into three separate summation equations.

n n n
E (i² ) + 3E (i) + E2 2
i=1 i=1 i=1

Then I'm not sure where to go from here(if that's even correct)...do you substitute the i=n(n+1)/2 formula in for the (i), i²=n(n+1)(2n+1)/6 for i² and the last summation is just 2n, then simplify? The answer I got from that was n³+3n²+2n+1, which is nothing even close to the answer that wolframalpha's calculator gave me.

Thanks for your help!

 

[Deleted member 4993]

Hello, got a question that I'm struggling with and can find no clear explanations on how to do it anywhere online:
Find the sum of:

n
E (i+1)(i+2)
i=1

(where E=sigma)

I figured that you should foil the brackets, making it i²+3i+2, then break that up into three separate summation equations.

n n n
E (i² ) + 3E (i) + E2 2
i=1 i=1 i=1

Then I'm not sure where to go from here(if that's even correct)...do you substitute the i= 3* n(n+1)/2 formula in for the (i), i²=n(n+1)(2n+1)/6 for i² and the last summation is just 2n, then simplify? The answer I got from that was n³+3n²+2n+1, which is nothing even close to the answer that wolframalpha's calculator gave me.

Thanks for your help!

Yes - except for the small correction as shown. If you want us to track down your mistake - you need to show us your process of simplification.

 

[xone306]

Ah, when I started typing it all out, I caught my mistake: I forgot to divide the first sum by six, I divided it all by two instead. Thanks for looking this over, I wasn't sure if I was even in the right ballpark!

Cheers!