Maclaurin series:

\(\displaystyle{f{{\left({x}\right)}}}={\sum_{{{n}={0}}}^{\infty}}{\frac{{{f}^{{{\left({n}\right)}{\left({0}\right)}}}}}{{{n}!}}}{x}^{{n}}={f{{\left({0}\right)}}}+{f}'{\left({0}\right)}{x}+{\frac{{{f}{''}{\left({0}\right)}}}{{{2}!}}}{x}^{{2}}+{\frac{{{f}{'''}{\left({0}\right)}}}{{{3}!}}}{x}^{{3}}+{\frac{{{{f}^{{{\left({4}\right)}}}{\left({0}\right)}}}}{{{4}!}}}{x}^{{4}}+\ldots\)

Here given \(f(0)=2,f'(0)=3,f''(0)=4\) and we need to find out the first 4 terms of the Maclaurin Series

\(\displaystyle{f{{\left({x}\right)}}}={\sum_{{{n}={0}}}^{\infty}}{\frac{{{{f}^{{{\left({n}\right)}}}{\left({0}\right)}}}}{{{n}!}}}{x}^{{n}}={f{{\left({0}\right)}}}+{f}'{\left({0}\right)}{x}+{\frac{{{f}{''}{\left({0}\right)}}}{{{2}!}}}{x}^{{2}}+{\frac{{{f}{'''}{\left({0}\right)}}}{{{3}!}}}{x}^{{3}}+\ldots\)

We need to find 4 terms

\(\displaystyle{f{{\left({x}\right)}}}={2}+{3}{x}+{\frac{{{4}}}{{{2}}}}{x}^{{2}}+{\frac{{{12}}}{{{6}}}}{x}^{{3}}\)

\(\displaystyle{f{{\left({x}\right)}}}={2}+{3}{x}+{2}{x}^{{2}}+{2}{x}^{{3}}\)

Result: \(f(x)=2+3x+2x^2+2x^3\)