### Question 11

- (i) Write down the Binomial expansion of (1 + x)4.

(ii) Use the result in **11**(a)(i) to evaluate, correct to three decimal places,

- The first, second and fifth terms of a linear sequence (A.P.) are three consecutive terms of an exponential sequence (G.P.). If the first term of the linear sequence is 7, find its common difference.

### Observation

This question was reported to attract majority of the candidates and they performed well in it. In part (a), while majority of them were able to write down the required expansion, a good number of them could not apply the expansion to solving the problem in part(a)(ii). Candidates were expected to apply the formula (1 + x)n = 1 + nx + nC2x2 + nC3x3 + … + xn to obtain (1 + x)4 = 1 + 4x + 6x2 + 4x3 + x4.

Now, ( = (1.25)4 = (1 + 0.25)4 = 1 + 4(0.25) + 6(0.25)2 + 4(0.25)3 + (0.25)4

= 1 + 1 + 0.375 + 0.0625 + 0.003906 = 2.441, correct to three decimal places.

In part (b), candidates were reported to apply the appropriate formulae correctly. Majority of them were able to recall the formula for the nth term of an Arithmetic Progression (AP) and that of geometric Progression (GP). Candidates were expected to show that if 7, 7 + d and 7 + 4d were the first, second and fifth terms of the AP respectively which were consecutive terms of the GP, then = . This implied that (7 + d)2 = 7(7 + 4d). Simplifying this equation gave d = 14.