O-LevelAdditional MathematicsSeries4037/22 · May/June 2023 · Q69 Marks

Find the first three terms in the expansion of (1 + x/7)⁵, in ascending powers of x. Simplify the coefficient of each term.

(a)(i)Find the first three terms in the expansion of (1 + x/7)⁵, in ascending powers of x. Simplify the coefficient of each term.[2]
(a)(ii)The expansion of 7(1+x)ⁿ(1 + x/7)⁵, where n is a positive integer, is written in ascending powers of x. The first two terms in the expansion are 7+89x. Find the value of n.[2]
(b)In the expansion of (k-2x)⁸, where k is a constant, the coefficient of x⁴ divided by the coefficient of x² is 5/8. The coefficient of x is positive. Form an equation and hence find the value of k.[5]

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

[B2] 1 + (5/7)x + (10/49)x²
[B1] for any two correct terms or for the three terms listed but not summed
[B2] 12
[B1] for 7n + 5 = 89 or 7(n + 5/7) = 89 oe or ⁿC₁ = 12
[B1] ⁸C₄k⁴(-2x)⁴ [x⁴] oe or 1120k⁴ [x⁴] soi
[B1] ⁸C₂k⁶(-2x)² [x²] oe or 112k⁶ [x²] soi
[M1] 1120k⁴/112k⁶ = 5/8 or 70×16×k⁴ / 28×4×k⁶ = 5/8 oe, soi FT providing at least B1 awarded and correct terms attempted
[A1] k² = 16 soi
[A1] [For coefficient of x to be positive k < 0, therefore] k = − 4

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.