Method of differences with given identity

2

1. Given that $$u _ { r } = \frac { 1 } { 6 } r ( r + 1 ) ( 4 r + 11 )$$ show that $$u _ { r } - u _ { r - 1 } = r ( 2 r + 3 )$$
2. Hence find the sum of the first hundred terms of the series $$1 \times 5 + 2 \times 7 + 3 \times 9 + \ldots + r ( 2 r + 3 ) + \ldots$$

2

1. Given that $$\frac { 1 } { r ( r + 1 ) ( r + 2 ) } = \frac { A } { r ( r + 1 ) } + \frac { B } { ( r + 1 ) ( r + 2 ) }$$ show that \(A = \frac { 1 } { 2 }\) and find the value of \(B\).
2. Use the method of differences to find $$\sum _ { r = 10 } ^ { 98 } \frac { 1 } { r ( r + 1 ) ( r + 2 ) }$$ giving your answer as a rational number.

4

1. Given that \(\mathrm { f } ( r ) = r ^ { 2 } \left( 2 r ^ { 2 } - 1 \right)\), show that $$\mathrm { f } ( r ) - \mathrm { f } ( r - 1 ) = ( 2 r - 1 ) ^ { 3 }$$
2. Use the method of differences to show that $$\sum _ { r = n + 1 } ^ { 2 n } ( 2 r - 1 ) ^ { 3 } = 3 n ^ { 2 } \left( 10 n ^ { 2 } - 1 \right)$$ (4 marks)

1

1. Given that \(\mathrm { f } ( r ) = \frac { 1 } { 4 r - 1 }\), show that $$\mathrm { f } ( r ) - \mathrm { f } ( r + 1 ) = \frac { A } { ( 4 r - 1 ) ( 4 r + 3 ) }$$ where \(A\) is an integer.
2. Use the method of differences to find the value of \(\sum _ { r = 1 } ^ { 50 } \frac { 1 } { ( 4 r - 1 ) ( 4 r + 3 ) }\), giving your answer as a fraction in its simplest form.
   [0pt] [4 marks]

1 In this question you must show detailed reasoning.
Find \(\sum _ { r = 1 } ^ { 100 } \left( \frac { 1 } { r } - \frac { 1 } { r + 2 } \right)\), giving your answer correct to 4 decimal places.

1 By expressing \(\frac { 1 } { r + 1 } - \frac { 1 } { r + 2 }\) as a single fraction, find \(\sum _ { r = 1 } ^ { n } \frac { 1 } { ( r + 1 ) ( r + 2 ) }\) in terms of \(n\).

9. (a) Given that \(A _ { r } = \frac { 1 } { r + 1 } - \frac { 2 } { r + 2 } + \frac { 1 } { r + 3 }\), show that \(A _ { r }\) can be expressed as \(\frac { 2 } { ( r + 1 ) ( r + 2 ) ( r + 3 ) }\).
(b) Hence, show that \(\sum _ { r = 1 } ^ { n } \frac { 2 } { ( r + 1 ) ( r + 2 ) ( r + 3 ) } = \frac { 1 } { 6 } - \frac { 1 } { ( n + 2 ) ( n + 3 ) }\).
(c) Find the ratio of \(\sum _ { r = 1 } ^ { 5 } A _ { r } : \sum _ { r = 1 } ^ { 10 } A _ { r }\), giving your answer in its simplest form.

1. Prove that

$$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( r + 1 ) ( r + 3 ) } = \frac { n ( a n + b ) } { 12 ( n + 2 ) ( n + 3 ) }$$ where \(a\) and \(b\) are constants to be found.

1. Use the method of differences to show that

$$\sum _ { r = 1 } ^ { n } \frac { 2 } { ( r + 4 ) ( r + 6 ) } = \frac { n ( a n + b ) } { 30 ( n + 5 ) ( n + 6 ) }$$ where \(a\) and \(b\) are integers to be determined.

1 In this question you must show detailed reasoning. Find \(\sum _ { r = 2 } ^ { 50 } \left( \frac { 1 } { r - 1 } - \frac { 1 } { r + 1 } \right)\), expressing the answer as an exact fraction.

17. (a) Express as a simplified fraction \(\frac { 1 } { ( r - 1 ) ^ { 2 } } - \frac { 1 } { r ^ { 2 } }\).
(2)
(b) Prove, by the method of differences, that $$\sum _ { r = 2 } ^ { n } \frac { 2 r - 1 } { r ^ { 2 } ( r - 1 ) ^ { 2 } } = 1 - \frac { 1 } { n ^ { 2 } }$$ [P4 June 2003 Qn 1]

1

1. Show that $$\frac { 1 } { r ^ { 2 } } - \frac { 1 } { ( r + 1 ) ^ { 2 } } = \frac { 2 r + 1 } { r ^ { 2 } ( r + 1 ) ^ { 2 } }$$
2. Hence find the sum of the first \(n\) terms of the series $$\frac { 3 } { 1 ^ { 2 } \times 2 ^ { 2 } } + \frac { 5 } { 2 ^ { 2 } \times 3 ^ { 2 } } + \frac { 7 } { 3 ^ { 2 } \times 4 ^ { 2 } } + \ldots$$

1

1. Given that $$\frac { r ^ { 2 } + r - 1 } { r ( r + 1 ) } = A + B \left( \frac { 1 } { r } - \frac { 1 } { r + 1 } \right)$$ find the values of \(A\) and \(B\).
2. Hence find the value of $$\sum _ { r = 1 } ^ { 99 } \frac { r ^ { 2 } + r - 1 } { r ( r + 1 ) }$$

7

1. Show that $$\frac { 1 } { r - 1 } - \frac { 1 } { r + 1 } \equiv \frac { A } { r ^ { 2 } - 1 }$$ where \(A\) is a constant to be found. 7
2. Hence use the method of differences to show that $$\sum _ { r = 2 } ^ { n } \frac { 1 } { r ^ { 2 } - 1 } \equiv \frac { a n ^ { 2 } + b n + c } { 4 n ( n + 1 ) }$$ where \(a\), \(b\) and \(c\) are integers to be found.