(a)     The sum of the first six terms of an arithmetic series is 81. The sum of its first eleven terms is 231. Find the first term and the common difference.

(b)     The sum of the first two terms of a geometric series is 1 and the sum of its first four terms is 5. If all of its terms are positive, find the first term and the common ratio.

(c)     The \({r^{{\text{th}}}}\) term of a new series is defined as the product of the \({r^{{\text{th}}}}\) term of the arithmetic series and the \({r^{{\text{th}}}}\) term of the geometric series above. Show that the \({r^{{\text{th}}}}\) term of this new series is \((r + 1){2^{r - 1}}\) .

Using mathematical induction, prove that

\[\sum\limits_{r = 1}^n {(r + 1){2^{r - 1}} = n{2^n},{\text{ }}n \in {\mathbb{Z}^ + }.} \]

(a)     \({S_6} = 81 \Rightarrow 81 = \frac{6}{2}(2a + 5d)\)     M1A1

\( \Rightarrow 27 = 2a + 5d\)

\({S_{11}} = 231 \Rightarrow 231 = \frac{{11}}{2}(2a + 10d)\)     M1A1

\( \Rightarrow 21 = a + 5d\)

solving simultaneously, a = 6 , d = 3     A1A1

[6 marks]

(b)     \(a + ar = 1\)     A1

\(a + ar + a{r^2} + a{r^3} = 5\)     A1

\( \Rightarrow (a + ar) + a{r^2}(1 + r) = 5\)

\( \Rightarrow 1 + a{r^2} \times \frac{1}{a} = 5\)

obtaining \({r^2} - 4 = 0\)     M1

\( \Rightarrow r = \pm 2\)

\(r = 2\,\,\,\,\,\)(since all terms are positive)     A1

\(a = \frac{1}{3}\)     A1

[5 marks]

(c)     \({\text{AP }}{r^{{\text{th}}}}{\text{ term is }}3r + 3\)     A1

\({\text{GP }}{r^{{\text{th}}}}{\text{ term is }}\frac{1}{3}{2^{r - 1}}\)     A1

\(3(r + 1) \times \frac{1}{3}{2^{r - 1}} = (r + 1){2^{r - 1}}\)     M1AG

[3 marks]

Total [14 marks]

prove: \({P_n}:\sum\limits_{r = 1}^n {(r + 1){2^{r - 1}} = n{2^n},{\text{ }}n \in {\mathbb{Z}^ + }.} \)

show true for n = 1 , i.e.

\({\text{LHS}} = 2 \times {2^0} = 2 = {\text{RHS}}\)     A1

assume true for n = k , i.e.     M1

\(\sum\limits_{r = 1}^k {(r + 1){2^{r - 1}} = k{2^k},{\text{ }}k \in {\mathbb{Z}^ + }} \)

consider n = k +1

\(\sum\limits_{r = 1}^{k + 1} {(r + 1){2^{r - 1}} = k{2^k} + (k + 2){2^k}} \)     M1A1

\( = {2^k}(k + k + 2)\)

\( = 2(k + 1){2^k}\)     A1

\( = (k + 1){2^{k + 1}}\)     A1

hence true for n = k + 1

\({P_{k + 1}}\) is true whenever \({P_k}\) is true, and \({P_1}\)is true, therefore \({P_n}\) is true     R1

for \(n \in {\mathbb{Z}^ + }\)

[7 marks]

Parts (a), (b) and (c) were answered successfully by a large number of candidates. Some, however, had difficulty with the arithmetic.

In part (d) many candidates showed little understanding of sigma notation and proof by induction. There were cases of circular reasoning and using n, k and r randomly. A concluding sentence almost always appeared, even if the proof was done incorrectly, or not done at all.