2014 H2 Mathematics Paper 1 Question 6

Sigma Notation

Answers

(ai)
Out of syllabus
(aii)
73n−49(4n−1){\frac{7}{3}n - \frac{4}{9}(4^n-1)}
(bi)
As n→∞,{n \to \infty, } 1(n+1)!→0{\frac{1}{(n+1)!} \to 0} so Sn→1{S_n \to 1}
Hence ∑ur{\sum u_r} converges
S∞=1{S_\infty = 1}
(bii)
un=n(n+1)!{u_n = \frac{n}{(n+1)!}}

Full solutions

(aii)
∑r=1npr=∑r=1n73−13∑r=1n4n=73n−13⋅4(4n−1)4−1=73n−49(4n−1)  ■\begin{align*} & \sum_{r=1}^{n} p_r \\ &= \sum_{r=1}^{n} \frac{7}{3} - \frac{1}{3} \sum_{r=1}^{n} 4^n \\ &= \frac{7}{3}n - \frac{1}{3} \cdot \frac{4(4^n-1)}{4-1} \\ &= \frac{7}{3}n - \frac{4}{9}(4^n-1) \; \blacksquare \end{align*}
(bi)
As n→∞,{n \to \infty, } 1(n+1)!→0{\displaystyle \frac{1}{(n+1)!} \to 0} so
Sn=1−1(n+1)!→1S_n = 1- \frac{1}{(n+1)!} \to 1
Hence ∑ur{\sum u_r} converges ■{\blacksquare}
S∞=1  ■S_\infty = 1 \; \blacksquare
(bii)
un=Sn−Sn−1=1−1(n+1)!−(1−1n!)=1n!−1(n+1)!=n+1(n+1)!−1(n+1)!=n(n+1)!  ■\begin{align*} u_n &= S_n - S_{n-1} \\ &= 1 - \frac{1}{(n+1)!} - \left( 1 - \frac{1}{n!} \right) \\ &= \frac{1}{n!} - \frac{1}{(n+1)!} \\ &= \frac{n+1}{(n+1)!} - \frac{1}{(n+1)!} \\ &= \frac{n}{(n+1)!} \; \blacksquare \end{align*}