Question

Using binomial theorem, evaluate ( 1.025) to the power of -1 / 3. Correct to four places of decimals?

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{series}\mathrm{which}\mathrm{arises}\mathrm{in}\mathrm{the}\mathrm{negative}\mathrm{binomial}\mathrm{theorem}\mathrm{for}\mathrm{negative}\mathrm{integer}-\mathrm{n} \\ (\mathrm{a}+\mathrm{x}{)}^{-\mathrm{n}}=\sum _{\mathrm{k}=0}^{\infty }{}^{-\mathrm{n}}\mathrm{C}_{\mathrm{k}}{\mathrm{a}}^{-\mathrm{n}-\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}=\sum _{\mathrm{k}=0}^{\infty }(-1{)}^{\mathrm{k}}\times {}^{\mathrm{n}+\mathrm{k}-1}\mathrm{C}_{\mathrm{k}}{\mathrm{a}}^{-\mathrm{n}-\mathrm{k}}{\mathrm{x}}^{\mathrm{k}} \\ \mathrm{For}\mathrm{a}=1 \\ (1+\mathrm{x}{)}^{-\mathrm{n}}=1-\mathrm{nx}+\frac{1}{2}\mathrm{n}(\mathrm{n}+1){\mathrm{x}}^2-\frac{1}{6}\mathrm{n}(\mathrm{n}+1\left)\right(\mathrm{n}+2){\mathrm{x}}^3+.... \\ \mathrm{Now} \\ (1.025{)}^{\frac{-1}{3}}=(1+.025{)}^{\frac{-1}{3}} \\ (1+.025{)}^{\frac{-1}{3}}=1-\frac{1}{3}\times 0.025+\frac{1}{2}\times \frac{1}{3}(\frac{1}{3}+1)(0.025{)}^2-\frac{1}{6}\times \frac{1}{3}(\frac{1}{3}+1\left)\right(\frac{1}{3}+2\left)\right(0.025{)}^3 \\ \mathrm{Neglecting}\mathrm{the}\mathrm{higher}\mathrm{terms} \\ (1+.025{)}^{\frac{-1}{3}}=0.9918 \end{gathered}$$