Alevel_C1C2
Alevel_C1C2
Alevel_C1C2
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13<br />
Exam-style mark scheme<br />
Sequences and series<br />
Question Scheme Marks<br />
Number<br />
1. (a) 4, 2, 1, 1 2<br />
A1 (1)<br />
(b)<br />
10<br />
41 ( − ( 05 . ) )<br />
S 10 = = 7. 9922 (4 dp.)<br />
1−<br />
0.<br />
5<br />
M1 A1 (2)<br />
(c) S ∞ =<br />
2<br />
= 8<br />
1−<br />
0.<br />
5<br />
M1 A1<br />
S∞ S10 0. 0078<br />
A1 (3)<br />
6<br />
2. (a) d = 5 ÷<br />
5<br />
− =<br />
3<br />
−<br />
1<br />
3<br />
M1 A1 (2)<br />
(b)<br />
5<br />
,<br />
5<br />
− A2 (2)<br />
81 243<br />
( )<br />
−<br />
(c) S ∞ = 5 ÷ 1 −<br />
1<br />
=<br />
3<br />
15<br />
4<br />
2 3 4<br />
or 3.75 M1 A1 (2)<br />
6<br />
3. (a) 16 + 32x + 24x + 8x + x A3 (3)<br />
2 3 4 2<br />
(b) ( 1 + 2x)( 16 + 32x + 24x + 8x + x ) ≈ 16 + 64x + 88x<br />
a = 16, b = 64, c = 88 A3 (3)<br />
6<br />
−<br />
( ) n<br />
( )<br />
2<br />
1<br />
4. (a) 1− = 1− × + ( )<br />
n<br />
× −<br />
2 2 2 2<br />
+ ... M1 A1<br />
nn ( − 1)<br />
∴ = 9<br />
8<br />
B1<br />
2<br />
n − n − 72 = 0<br />
B1<br />
( n − 9)( n + 8)<br />
= 0<br />
Since n is a positive, therefore n = 9 A1 (5)<br />
(b) Coefficient of x 3 : nn ( 1 )( n 2 )<br />
6<br />
5. (a)<br />
− − ×<br />
1<br />
( − )<br />
2<br />
3<br />
=<br />
−<br />
21<br />
2<br />
M1 A1 (2)<br />
8 × 7 2<br />
a = 448, a =±4 M1 A2 (3)<br />
2<br />
2 2 3 3<br />
(b) 1 + 8ax + 28a x + 56a x + ... M3<br />
Let x = 0.1, then<br />
B1<br />
8<br />
( 1 + 4 × 0. 1) ≈ 1 + 32 × 0. 1 + 448 × 0. 01 + 3584 × 0.<br />
001 = 12.264 M1 A1 (6)<br />
8<br />
(c) Percentage error: (.) 1 4 − 12 . 264<br />
× 100 = 17%<br />
8<br />
(nearest integer)<br />
(.) 14<br />
M1 A1 (2)<br />
11<br />
7<br />
© Oxford University Press 2008<br />
Core C2