2017 H2 Mathematics Paper 1 Question 9

Sigma Notation

Answers

(ai)
un=2An−A+B{u_n = 2An - A + B}
(aii)
A=3,  B=−9{A=3, \; B=-9}

(b)

k=4{k=4}
n2(n+1)24{\frac{n^2(n+1)^2}{4}}

(c)

ex{\mathrm{e}^x}

Full solutions

(ai)
un=Sn−Sn−1=An2+Bn−(A(n−1)2+B(n−1))=An2+Bn−(A(n2−2n+1)+B(n−1))=2An−A+B  ■\begin{align*} u_n &= S_n - S_{n-1} \\ &= An^2 + Bn - \left( A(n-1)^2 + B(n-1) \right) \\ &= An^2 + Bn - \left( A(n^2 - 2n + 1) + B(n-1) \right) \\ &= 2An - A + B \; \blacksquare \end{align*}
(aii)
Given u10=48{u_{10}=48} and u17=90{u_{17}=90},
19A+B=4833A+B=90\begin{align} && \quad 19A + B &= 48 \\ && \quad 33A + B &= 90 \end{align}
Solving with a GC,
A=3,  B=−9  ■A=3, \; B=-9 \; \blacksquare

(b)

r2(r+1)2−(r−1)2r2=r2((r+1)2−(r−1)2)=r2(r+1+r−1)(r+1−(r−1))=4r3  ■\begin{align*} & r^2(r+1)^2 - (r-1)^2r^2 \\ &= r^2\Big( (r+1)^2 - (r-1)^2 \Big) \\ &= r^2 (r+1+r-1)\Big( r+1-(r-1) \Big) \\ &= 4r^3 \; \blacksquare \end{align*}
∑r=1nr3=∑r=1nr2(r+1)2−(r−1)2r24=14(12⋅22−02⋅12+22⋅32−12⋅22+32⋅42−22⋅32⋯(n−2)2(n−1)2−(n−3)2(n−2)2(n−1)2n2−(n−2)2(n−1)2n2(n+1)2−(n−1)2n2)=n2(n+1)24  ■\begin{align*} & \sum_{r=1}^n r^3 \\ & = \sum_{r=1}^n \frac{r^2(r+1)^2 - (r-1)^2r^2}{4} \\ & = \def\arraystretch{1.5} \frac{1}{4} \left( \begin{array}{lclc} & \bcancel{1^2 \cdot 2^2} &-& 0^2 \cdot 1^2 \\ + & \bcancel{2^2 \cdot 3^2} &-& \bcancel{1^2 \cdot 2^2} \\ + & \bcancel{3^2 \cdot 4^2} &-& \bcancel{2^2 \cdot 3^2} \\ & & \cdots & \\ & \bcancel{(n-2)^2(n-1)^2} &-& \bcancel{(n-3)^2(n-2)^2} \\ & \bcancel{(n-1)^2n^2} &-& \bcancel{(n-2)^2(n-1)^2} \\ & n^2(n+1)^2 &-& \bcancel{(n-1)^2n^2} \\ \end{array} \right) \\ &= \frac{n^2 (n+1)^2}{4} \; \blacksquare \end{align*}

(c)

ar=xrr!a_r = \frac{x^r}{r!}
∣an+1an∣=∣xn+1(n+1)!÷xnn!∣=∣x∣n+1\begin{align*} \left| \frac{a_{n+1}}{a_n} \right| &= \left| \frac{x^{n+1}}{(n+1)!} \div \frac{x^n}{n!} \right| \\ &= \frac{\left| x \right|}{n+1} \end{align*}
As n→∞,{n \to \infty, } ∣x∣n+1→0{\frac{\left|x\right|}{n+1} \to 0}
lim⁡n→∞∣an+1an∣=0<1\lim_{n\to \infty} \left| \frac{a_{n+1}}{a_n} \right| = 0 < 1
Hence by D'Alembert's ratio test, ∑r=0∞xrr!{\displaystyle \sum_{r=0}^\infty \frac{x^r}{r!}} converges for all real values of x  ■{x \; \blacksquare}
∑r=0∞xrr!=1+x+x22!+x33!+…\sum_{r=0}^\infty \frac{x^r}{r!} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots
This is the Maclaurin series for ex{\mathrm{e}^x}
Hence the sum to infinity is ex  ■{\mathrm{e}^x \; \blacksquare}