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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

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