Given function is \(f(x)=(1+x)^{\frac{2}{3}}\)

For function \(f(x)=(1+x)^n\) binomial series centered at 0 is given as:

\((1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+...\)

We have, \(f(x)=(1+x)^{\frac{2}{3}}\)

Here, \(n=\frac{2}{3}\)

Putting \(n=\frac{2}{3}\) in the series, we get

\((1+x)^{\frac{2}{3}}=1+\frac{2}{3}x+\frac{\frac{2}{3}(\frac{2}{3}-1)}{2}x^2+\frac{\frac{2}{3}(\frac{2}{3}-1)(\frac{2}{3}-2)}{6}x^3+...\)

\(=1+\frac{2}{3}x+\frac{\frac{2}{3}\cdot\frac{-1}{3}}{2}x^2+\frac{\frac{2}{3}\cdot\frac{-1}{3}\cdot\frac{-4}{3}}{6}x^3+...\)

\(=1+\frac{2}{3}x-\frac{1}{9}x^2+\frac{4}{81}x^3+...\)

First four non-zero terms are:

\((1+x)^{\frac{2}{3}}=1+\frac{2}{3}x-\frac{1}{9}x^2+\frac{4}{81}x^3\)

Now, to approximate \((1.02)^\frac{2}{3}\)

\((1.02)^{\frac{2}{3}}=(1+0.02)^{\frac{2}{3}}\)

Put x=0.02 in the series we obtained above.

\((1+0.02)^{\frac{2}{3}}=1+\frac{2}{3}\cdot(0.02)-\frac{1}{9}\cdot(0.02)^2+\frac{4}{81}\cdot(0.02)^3\)

\(=1+\frac{0.04}{3}-\frac{0.0004}{9}+\frac{4}{81}\cdot(0.000008)\)

\(=1+0.0133-0.000044+0.000000392\)

\(=1.0128\)

For function \(f(x)=(1+x)^n\) binomial series centered at 0 is given as:

\((1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+...\)

We have, \(f(x)=(1+x)^{\frac{2}{3}}\)

Here, \(n=\frac{2}{3}\)

Putting \(n=\frac{2}{3}\) in the series, we get

\((1+x)^{\frac{2}{3}}=1+\frac{2}{3}x+\frac{\frac{2}{3}(\frac{2}{3}-1)}{2}x^2+\frac{\frac{2}{3}(\frac{2}{3}-1)(\frac{2}{3}-2)}{6}x^3+...\)

\(=1+\frac{2}{3}x+\frac{\frac{2}{3}\cdot\frac{-1}{3}}{2}x^2+\frac{\frac{2}{3}\cdot\frac{-1}{3}\cdot\frac{-4}{3}}{6}x^3+...\)

\(=1+\frac{2}{3}x-\frac{1}{9}x^2+\frac{4}{81}x^3+...\)

First four non-zero terms are:

\((1+x)^{\frac{2}{3}}=1+\frac{2}{3}x-\frac{1}{9}x^2+\frac{4}{81}x^3\)

Now, to approximate \((1.02)^\frac{2}{3}\)

\((1.02)^{\frac{2}{3}}=(1+0.02)^{\frac{2}{3}}\)

Put x=0.02 in the series we obtained above.

\((1+0.02)^{\frac{2}{3}}=1+\frac{2}{3}\cdot(0.02)-\frac{1}{9}\cdot(0.02)^2+\frac{4}{81}\cdot(0.02)^3\)

\(=1+\frac{0.04}{3}-\frac{0.0004}{9}+\frac{4}{81}\cdot(0.000008)\)

\(=1+0.0133-0.000044+0.000000392\)

\(=1.0128\)