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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/migrationdocuments/QP GCE Curriculum 2000/June 2013 - QP/6667_01R_que_20130610.pdf

Mark scheme: http://www.edexcel.com/migrationdocuments/QP GCE Curriculum 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.

Paper: http://www.edexcel.com/migrationdocuments/QP GCE Curriculum 2000/June 2013 - QP/6667_01R_que_20130610.pdf

Mark scheme: http://www.edexcel.com/migrationdocuments/QP GCE Curriculum 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.

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