# FP1 Series help

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#1
Use the standard results for n∑r=1 (r²) and n∑r=1 (r) to show that

[n∑r=0] (r²-2r+2n+1)=(1/6)(n+1)(n+a)(bn+c) for all integers n≥0, where a, b and c are constant integers to be found.

I’ve written the following so far:
[n∑r=0] =[n∑r=1] - [1∑r=1]

[n∑r=1] (r²-2r+2n+1)
=(n/6)(n+1)(2n+1)-2(n/2)(n+1)+2n+n
=(n/6)(n+1)(2n+1)-n(n+1)+3n
=(n/6)(n+1)(2n+1)-n²+2n

[1∑r=1] (r²-2r+2n+1)
=(1/6)(1+1)(2+1)-1(1+1)+3(1)
=2

⇒[n∑r=0] (r²-2r+2n+1) =(n/6)(n+1)(2n+1)-n²+2n+2

What do I do now, since it won’t factorise?
1
7 years ago
#2
(Original post by bobbricks)
I’ve written the following so far:
[n∑r=0] =[n∑r=1] - [1∑r=1]
I think it should be: [n∑r=0] =[n∑r=1] + [0∑r=0]

which is equal to: [n∑r=0] =[n∑r=1] + (0)^2 -2(0) + 2n +1

[n∑r=0] =[n∑r=1] + 2n +1
0
#3
(Original post by SherlockHolmes)
I think it should be: [n∑r=0] =[n∑r=1] + [0∑r=0]

which is equal to: [n∑r=0] =[n∑r=1] + (0)^2 -2(0) + 2n +1

[n∑r=0] =[n∑r=1] + 2n +1
hmmm...why would it be that (compared to my version)?
0
7 years ago
#4
(Original post by bobbricks)
hmmm...why would it be that (compared to my version)?

[n∑r=0] =[n∑r=1] - [1∑r=1]

You have said the sum from r=0 to n is equal to the sum of r=1 to n minus the value when r=1. This is incorrect because [n∑r=1] - [1∑r=1] is actually [n∑r=2].

My version:
[n∑r=0] =[n∑r=1] + [0∑r=0]

I have said the sum from r=0 to n is equal to the sum of r=1 to n plus the value when r=0.
0
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