What Is the Explanation for the 2n(n+1) in the FP1 Mark Scheme?

[bobbricks]

Can someone explain the second step in the mark scheme for 10ii? I don't understand where the 2n(n+1) and (n+1) comes from and what the meaning of having r and n terms is when they're both on the left hand side (I'm only used to seeing only r terms on the left hand side with the sigma sign)?

Paper: http://www.edexcel.com/migrationdoc...2000/June 2013 - QP/6667_01R_que_20130610.pdf
Mark scheme: http://www.edexcel.com/migrationdoc...2000/June 2013 - MS/6667_01R_msc_20130815.pdf

I wrote out the following:
[nΣr=0 (r²-2r+2n+1)] =[nΣr=1(r²-2r+2n+1)] +[0Σr=1(r²-2r+2n+1)]
which leads to (1/6) (n+1)(2n²-5n+12), which cannot be factorised further.
Then I thought, it could be [nΣr=0] (r²-2r+2n+1)] =[nΣr=1(r²-2r] +[0Σr=1(r²-2r]+2(n+1)+n which didn't work either, at which point I looked at the mark scheme.

 

[BvU]

I take it you accepted (otherwise check here)
$$\sum_{r=0}^n r^3 = {1\over 6} n(n+1)(2n+1) \ \ ;\ \ -2\sum_{r=0}^n r = -2{1\over 2} n(n+1)\ \ ;\ \ \sum_{r=0}^n (2n+1) = (n+1)(2n+1)$$The latter is rewritten as ##2n(n+1) + (n+1)##; then ##-n(n+1) + 2n(n+1) + (n+1) = (n+1)(n+1)##.

Factor out the ##{1\over 6}(n+1)## to get the ##{1\over 6}(n+1) \left \{ 2n^2+n \ + 6(n + 1) \right \}##

[edit] Can't check what went wrong with your "which leads to" unless you spell out how you obtained that.

 

[bobbricks]

I take it you accepted (otherwise check here)
$$\sum_{r=0}^n r^3 = {1\over 6} n(n+1)(2n+1) \ \ ;\ \ -2\sum_{r=0}^n r = -2{1\over 2} n(n+1)\ \ ;\ \ \sum_{r=0}^n (2n+1) = (n+1)(2n+1)$$The latter is rewritten as ##2n(n+1) + (n+1)##; then ##-n(n+1) + 2n(n+1) + (n+1) = (n+1)(n+1)##.

Factor out the ##{1\over 6}(n+1)## to get the ##{1\over 6}(n+1) \left \{ 2n^2+n \ + 6(n + 1) \right \}##

[edit] Can't check what went wrong with your "which leads to" unless you spell out how you obtained that.

That's the bit I don't understand- Why does the sum from r=0 to n of (2n+1)=(n+1)(2n+1)? -- I would've thought that [nΣr=0 (2n+1)] =[nΣr=0 (1)] +[nΣr=0(2n)] but then I'm not sure on what to do from there

 

[BvU]

You can take the (2n+1) out of the summation: ##\sum_{r=0}^n (2n+1) = (2n+1)\,\sum_{r=0}^n1## (the expression to be summed does not depend on ##r##!).

 

[bobbricks]

You can take the (2n+1) out of the summation: ##\sum_{r=0}^n (2n+1) = (2n+1)\,\sum_{r=0}^n1## (the expression to be summed does not depend on ##r##!).

So in order to find [nΣr=0 (2n+1)], would you have to do [nΣr=1(2n+1)]+[0Σr=1 (2n+1)] and I would've thought that would become [nΣr=0 (2n+1)]=[n(2n+1)]+[0(2n+1)]..?Also, how do you know (2n+1) is a constant?

 

[BvU]

You want to sum over r=0 to r = n. The only thing that changes from term to term is r.
Varying r doesn't change 2n+1

Terms are 2n+1, each and every one of them, for r=0 to r=n. So n+1 terms "2n+1". The sum is (n+1)(2n+1).
It is so evident, simple and so close in front of you that it might escape you for that reason only.

[edit] Tip: don't split sum r=0 to n in two unless you have a good reason.

 

[bobbricks]

You want to sum over r=0 to r = n. The only thing that changes from term to term is r.
Varying r doesn't change 2n+1

Terms are 2n+1, each and every one of them, for r=0 to r=n. So n+1 terms "2n+1". The sum is (n+1)(2n+1).
It is so evident, simple and so close in front of you that it might escape you for that reason only.

[edit] Tip: don't split sum r=0 to n in two unless you have a good reason.

Ah, I understand thanks :) ..I don't suppose you know where I could find more questions like this one just to see if I do truly know how to do it now since I don't have anything like this in my textbook, and have exams coming up.

 

[BvU]

Sorry, no. Not at home in that world.
And: good luck with the exams!

 

[bobbricks]

Sorry, no. Not at home in that world.
And: good luck with the exams!

Okay, thanks for the help though :)