Given a sequence of non negative integers \(\{ {a_r}\} \) show that

(i)     \(\sum\limits_{r = 0}^n {{a_r}{{(x + 1)}^r}(\bmod x) = \sum\limits_{r = 0}^n {{a_r}(\bmod x)} } \) where \(x \in \{ 2,{\text{ }}3,{\text{ }}4 \ldots \} \).

(ii)     \(\sum\limits_{r = 0}^n {(3{a_{2r + 1}} + {a_{2r}}){9^r} = \sum\limits_{r = 0}^{2n + 1} {{a_r}{3^r}} } \).

Hence determine whether the base \(3\) number \(22010112200201\) is divisible by \(8\).

(i)     \((x + 1)(\bmod x) \equiv 1(\bmod x)\)     (M1)

\({(x + 1)^r}(\bmod x)\left( { \equiv {1^r}(\bmod x)} \right) = 1(\bmod x)\)     A1

\(\sum\limits_{r - 0}^n {{a_r}{{(x + 1)}^r}(\bmod x) \equiv \sum\limits_{r = 0}^n {{a_r}(\bmod x)} } \)     AG

(ii)     METHOD 1

\(\sum\limits_{r = 0}^n {(3{a_{2r + 1}} + {a_{2r}}){9^r} = \sum\limits_{r = 0}^n {(3{a_{2r + 1}} + {a_{2r}}){3^{2r}}} } \)     M1

\( = \sum\limits_{r = 0}^n {({3^{2r + 1}}{a_{2r + 1}} + {3^{2r}}{a_{2r}})} \)     M1A1

\( = \sum\limits_{r = 0}^{2n + 1} {{a_r}{3^r}} \)     AG

METHOD 2

\(\sum\limits_{r = 0}^n {(3{a_{2r + 1}} + {a_{2r}}){9^r} = 3{a_1} + {a_0} + 9(3{a_3} + {a_2}) +  \ldots  + {9^n}(3{a_{2n + 1}} + {a_{2n}})} \)     A1

\( = {a_0} + 3{a_1} + {3^2}{a_2} + {3^3}{a_3} +  \ldots  + {3^{2n + 1}}{a_{2n + 1}}\)     M1A1

\( = \sum\limits_{r = 0}^{2n + 1} {{a_r}{3^r}} \)     AG

[5 marks]

METHOD 1

using part (a) (ii) to separate the number into pairs of digits     (M1)

\(22010112200201{\text{ }}({\text{base }}3) \equiv 8115621{\text{ }}({\text{base }}9)\)     A1

using the sum of digits identity from part (a) (i)     (M1)

Note:     Award M1 if result from a(i) is used to show that the number is divisible by \(2\), even if no other valid working given.

\({\text{sum}} = 24\)     A1

which is divisible by \(8\)     A1

hence \(22010112200201\) (base \(3\)) is divisible by \(8\)

METHOD 2

\(\sum\limits_{r = 0}^{13} {{a_r}{3^r}\sum\limits_{r = 0}^6 {(3{a_{2r + 1}} + {a_{2r}}){9^r}} } \)     (M1)

\( \equiv \sum\limits_{r = 0}^6 {(3{a_{2r + 1}} + {a_{2r}})(\bmod 8)} \)     A1

\( = {a_0} + 3{a_1} + {a_2} + 3{a_3} +  \ldots  + {a_{12}} + 3{a_{13}}(\bmod 8)\)     M1

\( = (1 + 3 \times 0 + 2 + 3 \times 0 +  \ldots  + 3 \times 2)(\bmod 8) \equiv 24(\bmod 8)\)     A1

\( \equiv 0\) OR divisible by \(8\)     A1

hence \(22010112200201\) (base \(3\)) is divisible by \(8\)

[5 marks]

Total [10 marks]