Edexcel A-Level Mathematics AS Fp1 7 Series Questions

$\sum_{r=1}^{n} r(r+1)(r-1)=\sum_{r=1}^{n} r^{3}-r$$=\frac{1}{4} n^{2}(n+1)^{2}-\frac{1}{2} n(n+1)$
(Please note the mark above is incorrectly labelled as A1 on e-PEN)

$=\frac{1}{4} n(n+1)\left[n^{2}+n-2\right]=\frac{1}{4} n(n+1)(n+\ldots)(n+\ldots)$

М1

$=\frac{1}{4} n(n+1)(n-1)(n+2)$
(4)

$$\sum_{r=n}^{2n}r^2 =\frac16(2n)(2n+1)(4n+1)-\frac16(n-1)n(2n-1)$$$$3\sum_{r=1}^{n}r(r+1)(r-1)=17\sum_{r=n}^{2n}r^2$$$$\frac34n(n+1)(n-1)(n+2) =\frac{17}{6}n\{2(8n^2+6n+1)-(2n^2-3n+1)\}$$$$18(n+1)(n-1)(n+2)=68(14n^2+15n+1)$$18(n+1)(n-1)(n+2)=68(14n+1)(n+1)18(n-1)(n+2)=68(14n+1)$$18n^2-934n-104=0$$n=52

$$\sum_{r=1}^{n}(r^2+2)=\sum_{r=1}^{n}r^2+\sum_{r=1}^{n}2$$$$=\frac{n}{6}(n+1)(2n+1)+2n$$$$=\frac{n}{6}(2n^2+3n+13)$$a=2, b=3, c=13

$$\sum_{r=10}^{25}(r^2+2)=S_{25}-S_9$$$$=\frac{25}{6}(2\cdot25^2+3\cdot25+13)-\frac{9}{6}(2\cdot9^2+3\cdot9+13)$$=5575-303=5272
Note: answer only from calculator is M0A0 as the question requires use of part (b).

$$\sum_{r=1}^{n}r(2r^2-3r-1) =\sum_{r=1}^{n}(2r^3-3r^2-r).$$$$=2\cdot\frac14n^2(n+1)^2 -3\cdot\frac16n(n+1)(2n+1) -\frac12n(n+1).$$$$=\frac12n(n+1)[n(n+1)-(2n+1)-1].$$$$=\frac12n(n+1)(n^2-n-2) =\frac12n(n+1)^2(n-2).$$
A1*

Using part (a),
$$\sum_{r=n}^{2n}r(2r^2-3r-1) =\sum_{r=1}^{2n}r(2r^2-3r-1) -\sum_{r=1}^{n-1}r(2r^2-3r-1).$$$$=\frac12(2n)(2n+1)^2(2n-2) -\frac12(n-1)n^2(n-3).$$$$=\frac12n(n-1)[4(2n+1)^2-n(n-3)].$$$$=\frac12n(n-1)(15n^2+19n+4).$$$$=\frac12n(n-1)(15n+4)(n+1).$$