Question
Find the value of k if \({\sum\limits_{r = 1}^\infty{k\left( {\frac{1}{3}} \right)}^r} = 7\).
Answer/Explanation
Markscheme
\({u_1} = \frac{1}{3}k{\text{ , }}r = \frac{1}{3}\) (A1) (A1)
\(7 = \frac{{\frac{1}{3}k}}{{1 – \frac{1}{3}}}\) M1
\(k = 14\) A1
[4 marks]
Examiners report
The question was well done generally. Those that did make mistakes on the question usually had the first term wrong, but did understand to use the formula for an infinite geometric series.
Question
a. Let \(\{ {u_n}\} ,{\text{ }}n \in {\mathbb{Z}^ + }\), be an arithmetic sequence with first term equal to \(a\) and common difference of \(d\), where \(d \ne 0\). Let another sequence \(\{ {v_n}\} ,{\text{ }}n \in {\mathbb{Z}^ + }\), be defined by \({v_n} = {2^{{u_n}}}\).
(i) Show that \(\frac{{{v_{n + 1}}}}{{{v_n}}}\) is a constant.
(ii) Write down the first term of the sequence \(\{ {v_n}\} \).
(iii) Write down a formula for \({v_n}\) in terms of \(a\), \(d\) and \(n\).[4]
(i) Find \({S_n}\), in terms of \(a\), \(d\) and \(n\).
(ii) Find the values of \(d\) for which \(\sum\limits_{i = 1}^\infty {{v_i}} \) exists.
You are now told that \(\sum\limits_{i = 1}^\infty {{v_i}} \) does exist and is denoted by \({S_\infty }\).
(iii) Write down \({S_\infty }\) in terms of \(a\) and \(d\) .
(iv) Given that \({S_\infty } = {2^{a + 1}}\) find the value of \(d\) .[8]
Find \(\sum\limits_{i = 1}^n {{z_i}} \) giving your answer in the form \(\ln k\) with \(k\) in terms of \(n\), \(p\) and \(q\).[6]
Answer/Explanation
Markscheme
a.
(i) METHOD 1
\(\frac{{{v_{n + 1}}}}{{{v_n}}} = \frac{{{2^{{u_{n + 1}}}}}}{{{2^{{u_n}}}}}\) M1
\( = {2^{{u_{n + 1}} – {u_n}}} = {2^d}\) A1
METHOD 2
\(\frac{{{v_{n + 1}}}}{{{v_n}}} = \frac{{{2^{a + nd}}}}{{{2^{a + (n – 1)d}}}}\) M1
\( = {2^d}\) A1
(ii) \( = {2^a}\) A1
Note: Accept \( = {2^{{u_1}}}\).
(iii) EITHER
\({v_n}\) is a GP with first term \({2^a}\) and common ratio \({2^d}\)
\({v_n} = {2^a}{({2^d})^{(n – 1)}}\)
OR
\({u_n} = a + (n – 1)d\) as it is an AP
THEN
\({v_n} = {2^a}^{ + (n – 1)d}\) A1
[4 marks]
(i) \({S_n} = \frac{{{2^a}\left( {{{({2^d})}^n} – 1} \right)}}{{{2^d} – 1}} = \frac{{{2^a}({2^{dn}} – 1)}}{{{2^d} – 1}}\) M1A1
Note: Accept either expression.
(ii) for sum to infinity to exist need \( – 1 < {2^d} < 1\) R1
\( \Rightarrow \log {2^d} < 0 \Rightarrow d\log 2 < 0 \Rightarrow d < 0\) (M1)A1
Note: Also allow graph of \({2^d}\).
(iii) \({S_\infty } = \frac{{{2^a}}}{{1 – {2^d}}}\) A1
(iv) \(\frac{{{2^a}}}{{1 – {2^d}}} = {2^{a + 1}} \Rightarrow \frac{1}{{1 – {2^d}}} = 2\) M1
\( \Rightarrow 1 = 2 – {2^{d + 1}} \Rightarrow {2^{d + 1}} = 1\)
\( \Rightarrow d = – 1\) A1
[8 marks]
METHOD 1
\({w_n} = p{q^{n – 1}},{\text{ }}{z_n} = \ln p{q^{n – 1}}\) (A1)
\({z_n} = \ln p + (n – 1)\ln q\) M1A1
\({z_{n + 1}} – {z_n} = (\ln p + n\ln q) – (\ln p + (n – 1)\ln q) = \ln q\)
which is a constant so this is an AP
(with first term \(\ln p\) and common difference \(\ln q\))
\(\sum\limits_{i = 1}^n {{z_i} = \frac{n}{2}\left( {2\ln p + (n – 1)\ln q} \right)} \) M1
\( = n\left( {\ln p + \ln {q^{\left( {\frac{{n – 1}}{2}} \right)}}} \right) = n\ln \left( {p{q^{\left( {\frac{{n – 1}}{2}} \right)}}} \right)\) (M1)
\( = \ln \left( {{p^n}{q^{\frac{{n(n – 1)}}{2}}}} \right)\) A1
METHOD 2
\(\sum\limits_{i = 1}^n {{z_i} = \ln p + \ln pq + \ln p{q^2} + \ldots + \ln p{q^{n – 1}}} \) (M1)A1
\( = \ln \left( {{p^n}{q^{\left( {1 + 2 + 3 + \ldots + (n – 1)} \right)}}} \right)\) (M1)A1
\( = \ln \left( {{p^n}{q^{\frac{{n(n – 1)}}{2}}}} \right)\) (M1)A1
[6 marks]
Total [18 marks]
Examiners report
a. Method of first part was fine but then some algebra mistakes often happened. The next two parts were generally good.
b. Given that (a) indicated that there was a common ratio a disappointing number thought it was an AP. Although some good answers in the next parts, there was also some poor notational misunderstanding with the sum to infinity still involving \(n\).