Edexcel IALGeneralGeneral Applications10 Marks
(a) Show that `\sum_{r=1}^{n} (8r - 3) = n(4n + 1)`. (b) Hence, or otherwise, find `n` such that `\sum_{r=1}^{n} (8r - 3) = 1170`. (c) Find `\sum_{r=10}^{20} (8r - 3)`.
- (a)Show that `\sum_{r=1}^{n} (8r - 3) = n(4n + 1)`.[3]
- (b)Hence, or otherwise, find `n` such that `\sum_{r=1}^{n} (8r - 3) = 1170`.[3]
- (c)Find `\sum_{r=10}^{20} (8r - 3)`.[4]
Mark scheme
Marking points from the official mark scheme. The mark or mark type is shown in brackets.
- M1 for identifying `a=5` (or `u_1=5`) and `d=8`.
- M1 for using the sum of an arithmetic series formula `S_n = n/2 (2a + (n-1)d)` or `S_n = n/2 (a + l)`.
- A1 for `n/2 (2(5) + (n-1)8) = n/2 (10 + 8n - 8) = n/2 (8n + 2) = n(4n + 1)` (cso).
- M1 for setting `n(4n+1) = 1170`.
- A1 for `4n^2 + n - 1170 = 0`.
- A1 for `n = 17` (rejecting negative solution).
- M1 for `S_{20} - S_9` OR `S_{20}` and `S_9` both explicitly stated.
- A1 for `S_{20} = 20(4(20)+1) = 20(81) = 1620`.
- A1 for `S_9 = 9(4(9)+1) = 9(37) = 333`.
- A1 for `1620 - 333 = 1287`.
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