(i)     Sum the series \(\sum\limits_{r = 0}^\infty  {{x^r}} \) .

(ii)     Hence, using sigma notation, deduce a series for

  (a)     \(\frac{1}{{1 + {x^2}}}\) ;

  (b)     \(\arctan x\) ;

  (c)     \(\frac{\pi }{6}\) .

Show that \(\sum\limits_{n = 1}^{100} {n! \equiv 3(\bmod 15)} \) .

(i)     \(\sum\limits_{r = 0}^\infty  {{x^r}}  = 1 + x + {x^2} + {x^3} + {x^4} +  \ldots  = \frac{1}{{1 - x}}\)     A1

(ii)     (a)     replacing x by \( - {x^2}\) gives     (M1)

\(\frac{1}{{1 - ( - {x^2})}} = 1 + ( - {x^2}) + {( - {x^2})^2} + {( - {x^2})^3} + {( - {x^2})^4} +  \ldots \)     A1

\(\frac{1}{{1 + {x^2}}} = 1 - {x^2} + {x^4} - {x^6} + {x^8} -  \ldots \)     (A1)

\( = \sum\limits_{r = 0}^\infty  {{{( - 1)}^r}{x^{2r}}} \)     A1     N2

(b)     \(\arctan x = \int {\frac{{{\rm{d}}x}}{{1 + {x^2}}}}  = x - \frac{{{x^3}}}{3} + \frac{{{x^5}}}{5} - \frac{{{x^7}}}{7} +  \ldots  + c\)     M1A1

\(x = 0 \Rightarrow c = 0\)     A1

\(\arctan x = \sum\limits_{r = 0}^\infty  {{{( - 1)}^r}\frac{{{x^{2r + 1}}}}{{2r + 1}}} \)     A1

(c)     by taking \(x = \frac{1}{{\sqrt 3 }}\)     M1

\(\arctan \frac{1}{{\sqrt 3 }} = \frac{\pi }{6} = \sum\limits_{r = 0}^\infty  {\frac{{{{( - 1)}^r}{{\left( {\frac{1}{{\sqrt 3 }}} \right)}^{2r + 1}}}}{{2r + 1}}} \)     A1

[11 marks]

\(\sum\limits_{n = 1}^{100} {n! = 1! + 2! + 3! + 4! + 5! +  \ldots } \)     M1     

\( = 1 + 2 + 6 + 24 + 120 +  \ldots \)

\( \equiv 1 + 2 + 6 + 24 + 0 + 0 + 0 +  \ldots (\bmod 15)\)     M1A1   

\( \equiv 33(\bmod 15)\)     A1

\( \equiv 3(\bmod 15)\)     AG

[4 marks]

In part (b) many did not recognize the sum of a simple geometric series to infinity and got involved in some heavy Maclaurin work thus wasting time. The clear instruction "Hence" was ignored by many candidates so that the question became more difficult and time consuming than it should have been.

Part (c) proved not to be as difficult as expected.