\[ \text{If we are prepared to take the assumption that }|x| < 1\\ \text{then we may resort to differentiation to obtain the desired answer.} \]
\[ \text{Note that differentiation of the infinite geometric series formula}\\ \text{(or any infinite series for that matter) is not well justified in the HSC}\\ \text{since there is no coverage on Taylor series at the HSC level.} \]
\[ \text{Nevertheless, the following computations are a classic way of forcing out}\\ \text{the end result. Note that due to the large number of derivatives}\\ \text{that would need to be typed up, I use computations from WolframAlpha.} \]
In practice, the quotient rule computations start getting a bit catastrophic after a while. Careful factorisation and use of chain rule is needed to avoid expanding nastily long binomial terms.
Note that whilst ridiculous for MX1, this is actually a fairly common technique in first year/second year university. It just so happens that we need to differentiate a lot.
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\[ \text{Begin by writing}\\ \frac{1}{1-x} = 1 + \sum_{n=1}^\infty x^n. \]
Note that the reason why the \(x^0\) term was pulled out is because it vanishes after the first derivative. Yet the following computations are rigged to ensure no other term vanishes.
\[ \text{Differentiating both sides gives}\\ \frac{1}{(1-x)^2} = \sum_{n=1}^\infty nx^{n-1}.\\ \text{But multiplying both sides now by }x\text{ gives}\\ \frac{x}{(1-x)^2} = \sum_{n=1}^\infty nx^n. \]
\[ \text{Differentiating both sides again now gives}\\ \frac{x+1}{(1-x)^3} = \sum_{n=1}^\infty n^2 x^{n-1}.\\ \text{Again multiplying both sides by }x\text{ gives}\\ \frac{x^2+x}{(1-x)^3} = \sum_{n=1}^\infty n^2 x^n. \]
Essentially the idea is to rinse and repeat.
\[ \text{Once again, differentiating,}\\ \frac{x^2+4x+1}{(1-x)^4} = \sum_{n=1}^\infty n^3 x^{n-1}.\\ \text{Multiplying }x\text{ to both sides yields}\\ \frac{x^3+4x^2+x}{(1-x)^4} = \sum_{n=1}^\infty n^3 x^n \]
\[ \text{Differentiating one last time gives}\\ \frac{x^3+11x^2+11x+1}{(1-x)^5} = \sum_{n=1}^\infty n^4 x^{n-1}. \]
\[ \text{Observe that the }RHS\text{ satisfies the criteria we require, i.e.}\\ \text{the }n\text{-th term of the sum, when expanded out,}\\ \text{will be }n^4 x^{n-1}. \]
\[ \text{Therefore equating coefficients,}\\ \begin{align*}a&=1\\ b&=11\\ c&=11\\ d&=1\\ e&=0 \end{align*}\]