(a) Write down the binomial expansion of \((2 - x)^{5}\) in ascending powers of x. (b) Use your expansion in (a) to evaluate \((1.98)^{5}\) correct to four ...

Answer Details

(a) Using \((a+b)^5=\sum_{k=0}^{5}\binom{5}{k}a^{5-k}b^{k}\) with \(a=2\) and \(b=-x\):

\[(2-x)^5=32-80x+80x^2-40x^3+10x^4-x^5\]

The coefficients come from \(\binom{5}{k}2^{5-k}(-1)^k\): \(32,\,-80,\,80,\,-40,\,10,\,-1\).

(b) Choose \(x\) so that \(2-x=1.98\), i.e. \(x=0.02\). Substitute:

• \(32\)
• \(-80(0.02)=-1.6\)
• \(+80(0.02)^2=+0.032\)
• \(-40(0.02)^3=-0.00032\)
• \(+10(0.02)^4=+0.0000016\)
• \(-(0.02)^5\approx 0\)

Adding: \(32-1.6+0.032-0.00032+0.0000016\approx 30.4316816\).

\[(1.98)^5\approx 30.4317\ \text{(4 d.p.)}\]