2020 H2 Mathematics Paper 2 Question 2

Sigma Notation

Answers

(A):

{\; 7, 9, 13, \ldots}

The sequence is increasing and diverges
(B):

{\; 5, 5, 5, \ldots}

The sequence is constant

{p=11}

{b=7}

{v_5 = 5a+28}

{u_n = 3 n^2 - 25 n + 16}

{m = 10}

(ai)

(A): If

{u_1 = p = 7, }

\begin{align*} u_2 &= 2(u_1)-5 \\ &= 2(7)-5 \\ &= 9 \\ u_3 &= 2(u_2)-5 \\ &= 2(9)-5 \\ &= 13 \\ \end{align*}

Hence the sequence is increasing and diverges

{\blacksquare}

(B): If

{u_1 = p = 5, }

\begin{align*} u_2 &= 2(u_1)-5 \\ &= 2(5)-5 \\ &= 5 \\ u_3 &= 2(u_2)-5 \\ &= 2(5)-5 \\ &= 5 \\ \end{align*}

Hence the sequence is constant

{\blacksquare}

(aii)

\begin{align*} u_5 &= 101 \\ 2u_4 - 5 &= 101 \\ u_4 &= 53 \\ 2u_3 - 5 &= 53 \\ u_3 &= 29 \\ 2u_2 - 5 &= 29 \\ u_2 &= 17 \\ 2u_1 - 5 &= 17 \\ p = u_1 &= 11 \; \blacksquare \end{align*}

(bi)

\begin{align*} u_4 &= 2v_3 \\ v_2 + 2 v_3 - 7 &= 2v_3 \\ b = v_2 &= 7 \; \blacksquare \end{align*}

(bii)

\begin{align*} v_5 &= v_3 + 2v_4 - 7 \\ &= v_3 + 2(2v_3) - 7 \\ &= 5v_3 - 7 \\ &= 5(v_1 + 2v_2 - 7) - 7 \\ &= 5(a+2(7)-7)-7 \\ &= 5a + 28 \; \blacksquare \end{align*}

(ci)

\begin{align*} u_n &= S_n - S_{n-1} \\ &= n^3 - 11n^2 + 4n - \Big( (n-1)^3 - 11(n-1)^2 + 4(n-1) \Big) \\ &= n^3 - 11n^2 + 4n - \Big( (n^3-3n^2+3n-1) - 11(n^2-2n+1) + 4(n-1) \Big) \\ &= 3 n^2 - 25 n + 16 \; \blacksquare \end{align*}

(cii)

\begin{gather*} S_m = S_3 \\ n^3 - 11 n^2 + 4 n = 3^3 - 11(3)^2+4(3) \\ n^3 - 11 n^2 + 4 n - 60 = 0 \\ (n - 3)(n + 2)(n - 10) = 0 \\ m=3 \textrm{ (NA)} \quad \textrm{or} \quad m=- 2 \textrm{ (NA)} \quad \textrm{or} \quad m=10 \\ m = 10 \; \blacksquare \end{gather*}