Specimen 2017 H2 Mathematics Paper 2 Question 2

Sigma Notation

Answers

{S_n = \frac{1}{3} n^3 - 2 n^2 + \frac{14}{3} n.}

{u_n = n^2 - 5 n + 7.}

{S_1 = 3,}

{S_2 = 4,}

{S_3 = 5}

and

{S_4 = 8.}

If ${S_n}$ is a quadratic polynomial in ${n,}$ ${S_n = an^2 + bn + c.}$

\begin{align} && \quad a + b + c &= 3 \\ && \quad 4a + 2b + c &= 4 \\ && \quad 9a + 3b + c &= 5 \\ && \quad 16a + 4b + c &= 8 \end{align}

Using a GC, there are no solutions to the above system of linear equations.

Hence ${S_n}$ cannot be a quadratic polynomial in ${n. \; \blacksquare}$

{S_n = an^3 + bn^2 + cn + d}

\begin{align} && \quad a + b + c + d &= 3 \\ && \quad 8a + 4b + 2c + d &= 4 \\ && \quad 27a + 9b + 3c + d &= 5 \\ && \quad 64a + 16b + 4c + d &= 8 \end{align}

Solving with a GC,

{a = \frac{1}{3},}

{b = - 2,}

{c = \frac{14}{3}}

and

{d = 0.}

S_n = \frac{1}{3} n^3 - 2 n^2 + \frac{14}{3} n \; \blacksquare

\begin{align*} S_{n-1} &= \frac{1}{3}(n-1)^3 - 2(n-1)^2 + \frac{14}{3}(n-1) \\ &= \frac{1}{3}(n^3 - 3 n^2 + 3 n - 1) - 2(n^2 - 2 n + 1) + \frac{14}{3}(n-1) \\ &= \frac{1}{3} n^3 - 3 n^2 + \frac{29}{3} n - 7 \end{align*}

\begin{align*} u_n &= S_n - S_{n-1} \\ &= \frac{1}{3} n^3 - 2 n^2 + \frac{14}{3} n - \left( \frac{1}{3} n^3 - 3 n^2 + \frac{29}{3} n - 7 \right)\\ &= n^2 - 5 n + 7 \; \blacksquare \end{align*}