(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}\...
Question 1 Report
(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.
Answer Details
(a)(i) Binomial expansion of \((1+x)^{4}\):
\[(1+x)^{4}=1+4x+6x^{2}+4x^{3}+x^{4}\](ii) Evaluate \(\left(\dfrac54\right)^{4}=\left(1+\dfrac14\right)^{4}\) with \(x=\dfrac14\):
\[=1+4\left(\tfrac14\right)+6\left(\tfrac14\right)^{2}+4\left(\tfrac14\right)^{3}+\left(\tfrac14\right)^{4}\]\[=1+1+0.375+0.0625+0.003906=2.441406\approx 2.441\](b) A.P. with first term 7; its 1st, 2nd and 5th terms form a G.P.
The three A.P. terms are \(7,\ 7+d,\ 7+4d\). For a G.P. the middle term squared equals the product of the outer terms:
\[(7+d)^{2}=7(7+4d)\]\[49+14d+d^{2}=49+28d\]\[d^{2}-14d=0\;\Rightarrow\;d(d-14)=0\]Taking the non-trivial value, the common difference is \(d=14\).
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