2016 H2 Mathematics Paper 1 Question 6

Sigma Notation

Answers

(i)

Out of syllabus

(ii)

u1=4,{u_1 = 4, } u2=14,{u_2 = 14, } u3=44{u_3 = 44}

(iii)

n4(n+1)(n2+n+2)+2{\frac{n}{4}(n+1)(n^2+n+2) + 2}

Full solutions

(i)

Out of syllabus

(ii)

u1=u0+13+1=2+1+1=4  ■\begin{align*} u_1 &= u_0 + 1^3 + 1 \\ &= 2 + 1 + 1 \\ &= 4 \; \blacksquare \end{align*}
u2=u1+23+2=4+8+2=14  ■\begin{align*} u_2 &= u_1 + 2^3 + 2 \\ &= 4 + 8 + 2 \\ &= 14 \; \blacksquare \end{align*}
u3=u2+33+3=14+27+3=44  ■\begin{align*} u_3 &= u_2 + 3^3 + 3 \\ &= 14 + 27 + 3 \\ &= 44 \; \blacksquare \end{align*}

(iii)

∑r=1n(ur−ur−1)=u1−u0+u2−u1+u3−u2+⋯+un−2−un−3+un−1−un−2+un−un−1=un−u0\begin{align*} &\sum_{r=1}^n (u_r - u_{r-1}) \\ & = \def\arraystretch{1.5} \begin{array}{lclc} & \bcancel{u_1} &-& u_0 \\ + & \bcancel{u_2} &-& \bcancel{u_1} \\ + & \bcancel{u_3} &-& \bcancel{u_2} \\ + & & \cdots & \\ + & \bcancel{u_{n-2}} &-& \bcancel{u_{n-3}} \\ + & \bcancel{u_{n-1}} &-& \bcancel{u_{n-2}} \\ + & u_n &-& \bcancel{u_{n-1}} \\ \end{array} \\ &= u_n - u_0 \end{align*}
∑r=1n(ur−ur−1)=un−2\begin{equation} \qquad \sum_{r=1}^{n} (u_r - u_{r-1}) = u_n - 2 \end{equation}
From parts (i) and (ii),
∑r=1nr(r2+1)=14n(n+1)(n2+n+2)\begin{equation} \qquad \sum_{r=1}^{n} r(r^2+1) = \frac{1}{4}n(n+1)(n^2+n+2) \end{equation}
un=un−1+n3+n\begin{equation} \qquad u_n = u_{n-1} + n^3 + n \end{equation}
Using (3){(3)} and (2),{(2),}
∑r=1n(ur−ur−1)=∑r=1n(r3+r)=∑r=1nr(r2+1)=14n(n+1)(n2+n+2)\begin{align*} &\sum_{r=1}^n (u_r - u_{r-1}) \\ &=\sum_{r=1}^n (r^3 + r) \\ &= \sum_{r=1}^n r(r^2 + 1) \\ &= \frac{1}{4}n(n+1)(n^2+n+2) \end{align*}
Using (1){(1)},
un−2=14n(n+1)(n2+n+2)un=14n(n+1)(n2+n+2)+2  ■\begin{align*} u_n - 2 &= \frac{1}{4}n(n+1)(n^2+n+2) \\ u_n &= \frac{1}{4}n(n+1)(n^2+n+2) + 2 \; \blacksquare \end{align*}