(i)     Express the sum of the first n positive odd integers using sigma notation.

(ii)     Show that the sum stated above is \({n^2}\).

(iii)     Deduce the value of the difference between the sum of the first 47 positive odd integers and the sum of the first 14 positive odd integers.

A number of distinct points are marked on the circumference of a circle, forming a polygon. Diagonals are drawn by joining all pairs of non-adjacent points.

(i)     Show on a diagram all diagonals if there are 5 points.

(ii)     Show that the number of diagonals is \(\frac{{n(n - 3)}}{2}\) if there are n points, where \(n > 2\).

(iii)     Given that there are more than one million diagonals, determine the least number of points for which this is possible.

The random variable \(X \sim B(n,{\text{ }}p)\) has mean 4 and variance 3.

(i)     Determine n and p.

(ii)     Find the probability that in a single experiment the outcome is 1 or 3.

(i)     \(\sum\limits_{k = 1}^n {(2k - 1)} \) (or equivalent)     A1

Note: Award A0 for \(\sum\limits_{n = 1}^n {(2n - 1)} \) or equivalent.

(ii)     EITHER

\(2 \times \frac{{n(n + 1)}}{2} - n\)     M1A1

OR

\(\frac{n}{2}\left( {2 + (n - 1)2} \right){\text{ (using }}{S_n} = \frac{n}{2}\left( {2{u_1} + (n - 1)d} \right))\)     M1A1

OR

\(\frac{n}{2}(1 + 2n - 1){\text{ (using }}{S_n} = \frac{n}{2}({u_1} + {u_n}))\)     M1A1

THEN

\( = {n^2}\)     AG

(iii)     \({47^2} - {14^2} = 2013\)     A1

[4 marks]

(i)     EITHER

a pentagon and five diagonals     A1

OR

five diagonals (circle optional)     A1

(ii)     Each point joins to n – 3 other points.     A1

a correct argument for \({n(n - 3)}\)     R1

a correct argument for \(\frac{{n(n - 3)}}{2}\)     R1

(iii)     attempting to solve \(\frac{1}{2}n(n - 3) > 1\,000\,000\) for n.     (M1)

\(n > 1415.7\)     (A1)

\(n = 1416\)     A1

[7 marks]

(i)     np = 4 and npq = 3     (A1)

attempting to solve for n and p     (M1)

\(n = 16\) and \(p = \frac{1}{4}\)     A1

(ii)     \(X \sim B(16,0.25)\)     (A1)

\(P(X = 1) = 0.0534538...( = \left( {\begin{array}{*{20}{c}}
  {16} \\
  1
\end{array}} \right)(0.25){(0.75)^{15}})\)     (A1)

\(P(X = 3) = 0.207876...( = \left( {\begin{array}{*{20}{c}}
  {16} \\
  3
\end{array}} \right){(0.25)^3}{(0.75)^{13}})\)     (A1)

\({\text{P}}(X = 1) + {\text{P}}(X = 3)\)     (M1)

= 0.261     A1

[8 marks]

In part (a) (i), a large number of candidates were unable to correctly use sigma notation to express the sum of the first n positive odd integers. Common errors included summing \(2n - 1\) from 1 to n and specifying sums with incorrect limits. Parts (a) (ii) and (iii) were generally well done.

Parts (b) (i) and (iii) were generally well done. In part (b) (iii), many candidates unnecessarily simplified their quadratic when direct GDC use could have been employed. A few candidates gave \(n > 1416\) as their final answer. While some candidates displayed sound reasoning in part (b) (ii), many candidates unfortunately adopted a ‘proof by example’ approach.

Part (c) was generally well done. In part (c) (ii), some candidates multiplied the two probabilities rather than adding the two probabilities.