2018 H2 Mathematics Paper 1 Question 8

Sigma Notation

Answers

(i)

A=5,  u3=40{A=5, \; u_3 = 40}

(ii)

a=7.5,  b=−5,  c=−5{a=7.5, \; b=-5, \; c=-5}

(iii)

15(2n−1)−5n(n+1)2−5n{15(2^n-1) - \frac{5n(n+1)}{2} - 5n}

Full solutions

(i)

u2=152u1+A=152(5)+A=15A=5  ■\begin{align*} u_2 &= 15 \\ 2u_1 + A &= 15 \\ 2(5) + A &= 15 \\ A &= 5 \; \blacksquare \end{align*}
u3=2u2+A(2)=2(15)+5(2)=40  ■\begin{align*} u_3 &= 2u_2 + A(2) \\ &= 2(15)+5(2) \\ &= 40 \; \blacksquare \end{align*}

(ii)

Using u1=5,{u_1 = 5,} u2=15,{u_2 = 15,} and u3=40,{u_3 = 40,}
2a+b+c=54a+2b+c=158a+3b+c=40\begin{align} && \quad 2a + b + c &= 5 \\ && \quad 4a + 2b + c &= 15 \\ && \quad 8a + 3b + c &= 40 \\ \end{align}
Solving with a GC,
a=7.5,  b=−5,  c=−5  ■a=7.5, \; b=-5, \; c=-5 \; \blacksquare

(iii)

∑r=1nur=7.5∑r=1n2r−5∑r=1nr−∑r=1n5=7.5⋅2(2n−1)2−1−5⋅n(n+1)2−5n=15(2n−1)−5n(n+1)2−5n  ■\begin{align*} &\sum_{r=1}^n u_r \\ &= 7.5 \sum_{r=1}^n 2^r -5 \sum_{r=1}^n r - \sum_{r=1}^n 5 \\ &= 7.5 \cdot \frac{2(2^n-1)}{2-1} - 5 \cdot \frac{n(n+1)}{2} - 5n \\ &= 15(2^n-1) - \frac{5n(n+1)}{2} - 5n \; \blacksquare \end{align*}