(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.