Problem source

\begin{questionparts}
\item Evaluate 
\[
\sum_{r=1}^{n}\frac{6}{r(r+1)(r+3)}.
\]
\item Expand $\ln(1+x+x^{2}+x^{3})$ as a series in powers of $x$,
where $\left|x\right|<1$, giving the first five non-zero terms and
the general term. 
\item Expand $\mathrm{e}^{x\ln(1+x)}$ as a series in powers of $x$,
where $-1 < x\leqslant1$, as far as the term in $x^{4}$. 
\end{questionparts}

Solution source

\begin{questionparts}
\item
\begin{align*}
&& \frac{6}{r(r+1)(r+3)} &= \frac{2}{r} - \frac{3}{r+1} + \frac{1}{r+3} \\
\Rightarrow && \sum_{r=1}^n \frac{6}{r(r+1)(r+3)} &= \sum_{r=1}^n \l \frac{2}{r} - \frac{3}{r+1} + \frac{1}{r+3} \r \\
&&&= \sum_{r=1}^n \frac{2}{r} - \sum_{r=1}^n \frac{3}{r+1} + \sum_{r=1}^n \frac{1}{r+3}  \\
&&&= \sum_{r=1}^n \frac{2}{r} - \sum_{r=2}^{n+1} \frac{3}{r} + \sum_{r=3}^{n+2} \frac{1}{r}  \\
&&& = \frac{2}{1} + \frac{2}{2} - \frac{3}{2} - \frac{3}{n+1}  + \frac{1}{n+1} + \frac{1}{n+2} \\
&&& = \frac{3}{2} - \frac{2}{n+1} + \frac{1}{n+2}
\end{align*}
\item \begin{align*}
&& \ln (1 + x+ x^2 + x^3) &= \ln \l \frac{1-x^4}{1-x} \r \\
&&&= \ln (1-x^4) - \ln(1-x) \\
&&&= \sum_{k=1}^{\infty} -\frac{x^{4k}}{k} - \sum_{k=1}^{\infty} - \frac{x^k}{k} \\
&&&= x + \frac12x^2+\frac13x^3-\frac34x^4+\frac15x^5 + \cdots \\
&&&= \sum_{k=1}^{\infty}a_k x^k 
\end{align*}
Where $a_k = \frac{1}{k}$ if $k \neq 0 \pmod{4}$ otherwise $a_k = -\frac{3}{k}$ if $k \equiv 0 \pmod{4}$
\item \begin{align*}
\exp(x \ln (1+x) ) &= \exp\l x \l x-\frac12x^2+\frac13x^3-\cdots \r \r \\
&= \exp\l x^2-\frac12x^3+\frac13x^4 \r \\
&= 1 + \l x^2-\frac12x^3+\frac13x^4 \r + \frac12 \l x^2-\frac12x^3+\frac13x^4 \r^2 + \cdots \\
&= 1 + x^2-\frac12x^3+\frac13x^4  + \frac12x^4 + \cdots \\
&= 1 + x^2 -\frac12x^3+\frac56x^4+\cdots
\end{align*}
\end{questionparts}