(a)(i) Write down the binomial expansion of ((1 + x)^{4}). (ii) Use the result in…

(a)(i) Write down the binomial expansion of ((1 + x)^{4}).

(ii) Use the result in (a)(i) to evaluate, correct to three decimal places ((frac{5}{4})^{4}).

(b) 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 the common difference.

Explanation

(a)(i) ((1 + x)^{4} = 1 + 4(x) + 6(x)^{2} + 4(x)^{3} + x^{4})

= (1 + 4x + 6x^{2} + 4x^{3} + x^{4})

(ii) ((frac{5}{4})^{4} = (1 + frac{1}{4})^{4})

= (1 + 4(frac{1}{4}) + 6(frac{1}{4})^{2} + 4(frac{1}{4})^{3} + (frac{1}{4})^{4})

= (1 + 1 + 0.375 + 0.0625 + 0.00390625)

= (2 + 0.44140625)

(approxeq 2.441)

(b) (T_{1} = a = 7) (first term of an A.P)

(T_{2} = a + d = 7 + d …. (1)) 

(T_{5} = a + 4d = 7 + 4d …. (2))

(T_{1} = 7 = a) (first term of G.P)

(T_{2} = ar = 7r)

(T_{3} = ar^{2} = 7r^{2}) 

(implies 7r = 7 + d)

(7r^{2} = 7 + 4d)

((1) times 4 : 28r = 28 + 4d … (3))

((3) – (2) : 28r – 7r^{2} = 28 – 7 = 21)

(7r^{2} – 28r + 21 = 0)

(7r^{2} – 21r – 7r + 21 = 0)

(7r(r – 3) – 7(r – 3) = 0)

((7r – 7)(r – 3) = 0)

(text{r = 1 or 3})

From (1), if r = 1,

(7 = 7 + d implies d = 0)  (This cannot be the case)

r = 3,

(7(3) = 7 + d implies d = 21 – 7 = 14)

The common difference = 14.