polynomial exercise asked by Katrina Linda and solved by using Geometric Series with its derivative

Exercise:

Let $f\left( x \right)=1+2x+3{{x}^{2}}+4{{x}^{3}}+...+98{{x}^{97}}+99{{x}^{98}}$

What is the value of $f\left( \frac{100}{99} \right)$

Solution: we have $f\left( x \right)=1+2x+3{{x}^{2}}+4{{x}^{3}}+...+99{{x}^{98}}=\sum\limits_{i=0}^{99}{i{{x}^{i-1}}}$

So $f\left( x \right)=\sum\limits_{i=0}^{99}{i{{x}^{i-1}}}=\frac{d}{dx}\left( \sum\limits_{i=0}^{99}{{{x}^{i}}} \right)$

But \(\sum\limits_{i=0}^{99}{{{x}^{i}}=}1+x+{{x}^{2}}+...+{{x}^{99}}\) is a Geometric series so $\sum\limits_{i=0}^{99}{{{x}^{i}}=\frac{{{x}^{100}}-1}{x-1}}$

Thus $\frac{d}{dx}\left( \sum\limits_{i=0}^{99}{{{x}^{i}}} \right)=\frac{d}{dx}\left( \frac{{{x}^{100}}-1}{x-1} \right)=\frac{100{{x}^{99}}\left( x-1 \right)-{{x}^{100}}+1}{{{\left( x-1 \right)}^{2}}}$

Thus $f\left( \frac{100}{99} \right)=\frac{100{{\left( \frac{100}{99} \right)}^{99}}\left( \frac{100}{99}-1 \right)-{{\left( \frac{100}{99} \right)}^{100}}+1}{{{\left( \frac{100}{99}-1 \right)}^{2}}}$

                  $=\frac{\frac{{{100}^{100}}}{{{99}^{99}}}\left( \frac{1}{99} \right)-\frac{{{100}^{100}}}{{{99}^{100}}}+1}{\frac{1}{{{99}^{2}}}}=\frac{1}{\frac{1}{{{99}^{2}}}}={{99}^{2}}=9801$

Or we can solve it by using $f\left( x \right)=\sum\limits_{i=0}^{98}{\left( i+1 \right){{x}^{i}}=\sum\limits_{i=0}^{98}{i{{x}^{i}}+\sum\limits_{i=0}^{98}{{{x}^{i}}}}}$

Notice that $\sum\limits_{i=0}^{98}{{{x}^{i}}=1+x+{{x}^{2}}+...+{{x}^{98}}}$ is Geometric Series hence $\sum\limits_{i=0}^{98}{{{x}^{i}}}=\frac{{{x}^{99}}-1}{x-1}$

let \(S=\sum\limits_{i=0}^{98}{i{{x}^{i}}}\Leftrightarrow S=\sum\limits_{i=0}^{98}{i{{x}^{i-1+1}}}=\sum\limits_{i=0}^{98}{\left( i{{x}^{i-1}} \right)x=x\frac{d}{dx}\left( \sum\limits_{i=0}^{98}{{{x}^{i}}} \right)}=x\frac{d}{dx}\left( \frac{{{x}^{99}}-1}{x-1} \right)\)

   $=\frac{x\left( 99{{x}^{98}}\left( x-1 \right)-{{x}^{99}}+1 \right)}{{{\left( x-1 \right)}^{2}}}=\frac{x\left( 99{{x}^{99}}-99{{x}^{98}}-{{x}^{99}}+1 \right)}{{{\left( x-1 \right)}^{2}}}=\frac{x\left( 98{{x}^{99}}-99{{x}^{98}}+1 \right)}{{{\left( x-1 \right)}^{2}}}$

So $f\left( x \right)=\frac{x\left( 98{{x}^{99}}-99{{x}^{98}}+1 \right)}{{{\left( x-1 \right)}^{2}}}+\frac{{{x}^{99}}-1}{x-1}=\frac{98{{x}^{100}}-99{{x}^{99}}+x+\left( x-1 \right)\left( {{x}^{99}}-1 \right)}{{{\left( x-1 \right)}^{2}}}$

            $=\frac{98{{x}^{100}}-99{{x}^{99}}+x+{{x}^{100}}-x-{{x}^{99}}+1}{{{\left( x-1 \right)}^{2}}}=\frac{99{{x}^{100}}-100{{x}^{99}}+1}{{{\left( x-1 \right)}^{2}}}$

Or we can use $u=i+1\Leftrightarrow i=u-1$

So $f\left( x \right)=\sum\limits_{i=0}^{98}{\left( i+1 \right){{x}^{i}}}=\sum\limits_{u=1}^{98}{u{{x}^{u-1}}}=\frac{d}{dx}\left( \sum\limits_{u=1}^{98}{{{x}^{u}}} \right)=\frac{d}{dx}\left( \frac{x\left( {{x}^{98}}-1 \right)}{x-1} \right)$