Top Notch

first, in summation form $\sum_{n=1}^\infty (\dfrac{(n-1)x^{n}}{n(n+1)})$ now, factor $\begin{array}{c}=&\dfrac{(n-1)x^{n}}{n(n+1)}\\ =&\dfrac{nx^{n}}{n(n+1)}-\dfrac{x^{n}}{n(n+1)}\\ =&\dfrac{x^{n}}{n+1}-x^n(\dfrac{1}{n(n+1)})\\ =&\dfrac{x^{n}}{n+1}-x^n(\dfrac{1}{n}-\dfrac{1}{n+1})\\ =&2\dfrac{x^{n}}{n+1}-\dfrac{x^n}{n} \end{array}$ hence $\sum_{n=1}^\infty (\dfrac{(n-1)x^{n}}{n(n+1)})=2\sum_{n=1}^\infty (\dfrac{x^{n}}{n+1})-\sum_{n=1}^\infty(\dfrac{x^n}{n})$ by a Maclaurin series expansions, $\sum_{n=1}^\infty(\dfrac{x^n}{n})=-\log (1-x)$ and $\begin{array}{c}=&\sum_{n=1}^\infty(\dfrac{x^n}{n+1})\\=&\dfrac{x}{2}+\dfrac{x^2}{3}+\dfrac{x^3}{4}......\\=&\dfrac{\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dfrac{x^4}{4}...}{x}\\=&\dfrac{x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dfrac{x^4}{4}.....-x}{x}\\=&\dfrac{\sum_{n=1}^\infty(\dfrac{x^n}{n})}{x}-1\\=&-\dfrac{\log (1-x)}{x}-1\end{array}$ hence $2\sum_{n=1}^\infty (\dfrac{x^{n}}{n+1})-\sum_{n=1}^\infty(\dfrac{x^n}{n})=2(-\dfrac{\log (1-x)}{x}-1)+\log (1-x)=(1-\dfrac{2}{x})\log (1-x)-2$ and $1^4+2^4=\boxed{17}$ . the question must mention that $|x|<1$ and i dont see what is so algebric 'bout it, more of a calculus problem.

Consider the series -

$1 + x + {x}^{2} + {x}^{3} + {x}^{4} +....... = \frac{1}{1-x}$

Differentiating and multiplying by $x$ ,

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

Integrating,

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

Using u-substitution as $u = 1 -x$ ,

$\frac{{x}^{2}}{2} + \frac{2{x}^{3}}{3} + .... = \int{\frac{1 - u}{{u}^{2}}}$

$=\int{\frac{1}{{u}^{2}}} - \int{\frac{1}{u}}$

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

Again integrating,

$\frac{{x}^{3}}{3.2} + \frac{2{x}^{4}}{4.3} + .... = -\int{\frac{1}{1-x}} - \int{ln(1-x)}$

Once again using u-substitution and IBP,

$\frac{{x}^{3}}{3.2} + \frac{2{x}^{4}}{4.3} + .... = ln(1-x) + (1-x)ln(1-x) + x$

$\frac{{x}^{3}}{3.2} + \frac{2{x}^{4}}{4.3} + .... = (2-x)ln(1-x) + x$

Dividing by x,

$\frac{{x}^{2}}{3.2} + \frac{2{x}^{3}}{4.3}+..... = (\frac{2}{x} - 1)ln(1-x) + 1$

${p}^{4} + {q}^{4} = {2}^{4} + {1}^{4} = 17$