Challenging +1.3 This is a structured proof by induction with a given hint that makes the algebraic manipulation straightforward. While it requires understanding of binomial coefficients, derivatives, and the product rule, the framework is standard and the Pascal's triangle identity is provided. More challenging than routine induction proofs due to the summation notation and derivative context, but less demanding than problems requiring novel insight or complex algebraic manipulation.
It is given that \(y = e^x u\), where \(u\) is a function of \(x\). The \(r\)th derivatives \(\frac{\mathrm{d}^r y}{\mathrm{d}x^r}\) and \(\frac{\mathrm{d}^r u}{\mathrm{d}x^r}\) are denoted by \(y^{(r)}\) and \(u^{(r)}\) respectively. Prove by mathematical induction that, for all positive integers \(n\),
$$y^{(n)} = e^x\left[\binom{n}{0}u + \binom{n}{1}u^{(1)} + \binom{n}{2}u^{(2)} + \ldots + \binom{n}{r}u^{(r)} + \ldots + \binom{n}{n}u^{(n)}\right].$$ [8]
[You may use without proof the result \(\binom{k}{r} + \binom{k}{r-1} = \binom{k+1}{r}\).]
It is given that $y = e^x u$, where $u$ is a function of $x$. The $r$th derivatives $\frac{\mathrm{d}^r y}{\mathrm{d}x^r}$ and $\frac{\mathrm{d}^r u}{\mathrm{d}x^r}$ are denoted by $y^{(r)}$ and $u^{(r)}$ respectively. Prove by mathematical induction that, for all positive integers $n$,
$$y^{(n)} = e^x\left[\binom{n}{0}u + \binom{n}{1}u^{(1)} + \binom{n}{2}u^{(2)} + \ldots + \binom{n}{r}u^{(r)} + \ldots + \binom{n}{n}u^{(n)}\right].$$ [8]
[You may use without proof the result $\binom{k}{r} + \binom{k}{r-1} = \binom{k+1}{r}$.]
\hfill \mbox{\textit{CAIE FP1 2018 Q6 [8]}}