21.2 Practice Problems
Problem 21.1. Find the \(8^{\text {th}}\) term in the expansion of \(\left (x^{2}+\frac {2}{x}\right )^{11}\).
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Solution. The general term is \[T_{r+1}=\binom {11}{r}\left (x^{2}\right )^{11-r}\left (\frac {2}{x}\right )^{r},\] so the \(8^{\text {th}}\) term has \(r=7\) — the index is always one less than the term number, since the first term is \(r=0\). Then \[T_{8}=\binom {11}{7}\left (x^{2}\right )^{4}\cdot \frac {2^{7}}{x^{7}} =330\cdot 128\cdot \frac {x^{8}}{x^{7}}=42240\,x .\]
Problem 21.2. Use the binomial series to expand \((1+x)^{\frac {1}{2}}\) as far as the term in \(x^{3}\), stating the values of \(x\) for which the expansion is valid. Hence find \(\sqrt {1.02}\) to five significant figures.
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Solution. With \(n=\frac {1}{2}\), \begin {align*} (1+x)^{1/2} &=1+\tfrac {1}{2}x +\frac {\tfrac {1}{2}\left (-\tfrac {1}{2}\right )}{2!}x^{2} +\frac {\tfrac {1}{2}\left (-\tfrac {1}{2}\right )\left (-\tfrac {3}{2}\right )}{3!}x^{3}+\cdots \\ &=1+\frac {x}{2}-\frac {x^{2}}{8}+\frac {x^{3}}{16}-\cdots , \end {align*}
valid for \(\left |x\right |<1\).
For \(\sqrt {1.02}\) take \(x=0.02\): \[1+0.01-\frac {0.0004}{8}+\frac {0.000008}{16} =1+0.01-0.00005+0.0000005=1.0099505 ,\] so \(\sqrt {1.02}\approx 1.0100\) to five significant figures.
Note 21.7. The approximation is good because \(x=0.02\) is small: each term is roughly a fiftieth of the one before, so three terms already fix the answer to seven decimal places. The true value is \(1.00995049\ldots \), and the sum above differs from it in the eighth.
That is the practical use of the series — not to expand \((1+x)^{1/2}\) symbolically, but to compute roots to whatever accuracy is wanted using only arithmetic. Before calculators this was how such values were found, and it is still how a calculator finds them.
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