Answer: (a)(i) \(1+21x+189x^2+945x^3\), (a)(ii) \(1.03^7=1.23\) to \(2\) d.p., (b) \(232960\).

Answer: (a)(i) \(1+21x+189x^2+945x^3\), (a)(ii) \(1.03^7=1.23\) to \(2\) d.p., (b) \(232960\).

(a)(i) By the binomial theorem,

\((1+3x)^7=1+7(3x)+\binom72(3x)^2+\binom73(3x)^3+\cdots.\)

So, up to the term in \(x^3\),

\((1+3x)^7=1+21x+21(9x^2)+35(27x^3)+\cdots.\)

Hence

\((1+3x)^7=1+21x+189x^2+945x^3+\cdots.\)

(a)(ii) To estimate \(1.03^7\), use \(1+3x=1.03\), so \(x=0.01\).

Then

\(1.03^7\approx1+21(0.01)+189(0.01)^2+945(0.01)^3.\)

Thus

\(1.03^7\approx1+0.21+0.0189+0.000945=1.229845.\)

Correct to \(2\) decimal places, this is \(1.23\).

(b) The general term in

\(\left(\frac{x^4}{2}+\frac{2}{x}\right)^{15}\)

is

\(\binom{15}{r}\left(\frac{x^4}{2}\right)^{15-r}\left(\frac{2}{x}\right)^r.\)

The power of \(x\) in this term is

\(4(15-r)-r=60-5r.\)

For the term independent of \(x\), set \(60-5r=0\), giving \(r=12\).

The required coefficient is therefore

\(\binom{15}{12}\left(\frac12\right)^3 2^{12}.\)

Since \(\binom{15}{12}=\binom{15}{3}=455\), this is

\(455\cdot2^9=455\cdot512=232960.\)