Method of differences with given identity

2

1. Use standard results from the List of formulae (MF19) to find \(\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } \left( 1 - \mathrm { r } - \mathrm { r } ^ { 2 } \right)\) in terms of \(n\),
   simplifying your answer. simplifying your answer.
2. Show that $$\frac { 1 - r - r ^ { 2 } } { \left( r ^ { 2 } + 2 r + 2 \right) \left( r ^ { 2 } + 1 \right) } = \frac { r + 1 } { ( r + 1 ) ^ { 2 } + 1 } - \frac { r } { r ^ { 2 } + 1 }$$ and hence use the method of differences to find \(\sum _ { r = 1 } ^ { n } \frac { 1 - r - r ^ { 2 } } { \left( r ^ { 2 } + 2 r + 2 \right) \left( r ^ { 2 } + 1 \right) }\).
3. Deduce the value of \(\sum _ { r = 1 } ^ { \infty } \frac { 1 - r - r ^ { 2 } } { \left( r ^ { 2 } + 2 r + 2 \right) \left( r ^ { 2 } + 1 \right) }\).

1 Let \(a\) be a positive constant.

1. Use the method of differences to find \(\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } \frac { 1 } { ( \mathrm { ar } + 1 ) ( \mathrm { ar } + \mathrm { a } + 1 ) }\) in terms of \(n\) and \(a\).
2. Find the value of \(a\) for which \(\sum _ { r = 1 } ^ { \infty } \frac { 1 } { ( a r + 1 ) ( a r + a + 1 ) } = \frac { 1 } { 6 }\).

1

1. Use standard results from the list of formulae (MF19) to find \(\sum _ { r = 1 } ^ { n } \left( 3 r ^ { 2 } + 3 r + 1 \right)\) in terms of \(n\), simplifying your answer.
2. Show that $$\frac { 1 } { r ^ { 3 } } - \frac { 1 } { ( r + 1 ) ^ { 3 } } = \frac { 3 r ^ { 2 } + 3 r + 1 } { r ^ { 3 } ( r + 1 ) ^ { 3 } }$$ and hence use the method of differences to find \(\sum _ { r = 1 } ^ { n } \frac { 3 r ^ { 2 } + 3 r + 1 } { r ^ { 3 } ( r + 1 ) ^ { 3 } }\).
3. Deduce the value of \(\sum _ { r = 1 } ^ { \infty } \frac { 3 r ^ { 2 } + 3 r + 1 } { r ^ { 3 } ( r + 1 ) ^ { 3 } }\).

1

1. Gie it h \(\mathrm { f } ( r ) = \frac { 1 } { ( r + 1 ) ( r + 2 ) } , \mathrm { s }\) th t $$\mathrm { f } \left( r - 1 \quad \mathrm { f } ( r ) = \frac { 2 } { r ( r + 1 ) ( r + 2 ) } \right.$$
2. Hen e fid \(\sum _ { r = 1 } ^ { n } \frac { 1 } { r ( r + 1 ) ( r + 2 ) }\).
3. Ded e th le \(6 \sum _ { r = 1 } ^ { \infty } \frac { 1 } { r ( r + 1 ) ( r + 2 ) }\).

2. (a) Show that, for \(r > 0\) $$\frac { r + 2 } { r ( r + 1 ) } - \frac { r + 3 } { ( r + 1 ) ( r + 2 ) } = \frac { r + 4 } { r ( r + 1 ) ( r + 2 ) }$$ (b) Hence show that $$\sum _ { r = 1 } ^ { n } \frac { r + 4 } { r ( r + 1 ) ( r + 2 ) } = \frac { n ( a n + b ) } { c ( n + 1 ) ( n + 2 ) }$$ where \(a\), \(b\) and \(c\) are integers to be determined.

1. (a) Express

$$\frac { 1 } { ( 2 n - 1 ) ( 2 n + 1 ) ( 2 n + 3 ) }$$ in partial fractions.
(b) Hence, using the method of differences, show that for all integer values of \(n\), $$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r - 1 ) ( 2 r + 1 ) ( 2 r + 3 ) } = \frac { n ( n + 2 ) } { a ( 2 n + b ) ( 2 n + c ) }$$ where \(a\), \(b\) and \(c\) are integers to be determined.

1. (a) Show that

$$\frac { 1 } { ( r + 1 ) ( r + 2 ) ( r + 3 ) } \equiv \frac { 1 } { 2 ( r + 1 ) ( r + 2 ) } - \frac { 1 } { 2 ( r + 2 ) ( r + 3 ) }$$ (b) Hence, or otherwise, find $$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( r + 1 ) ( r + 2 ) ( r + 3 ) }$$ giving your answer as a single fraction in its simplest form.

4. (a) Express as a simplified single fraction \(\frac { 1 } { ( r - 1 ) ^ { 2 } } - \frac { 1 } { r ^ { 2 } }\).
(b) Hence prove, by the method of differences, that \(\quad \sum _ { r = 2 } ^ { n } \frac { 2 r - 1 } { r ^ { 2 } ( r - 1 ) ^ { 2 } } = 1 - \frac { 1 } { n ^ { 2 } }\).

1. (a) Express \(\frac { 2 } { 4 r ^ { 2 } - 1 }\) in partial fractions.
   (b) Hence use the method of differences to show that

$$\sum _ { r = 1 } ^ { n } \frac { 1 } { 4 r ^ { 2 } - 1 } = \frac { n } { 2 n + 1 }$$

2. (a) Show that, for \(r > 0\) $$r - 3 + \frac { 1 } { r + 1 } - \frac { 1 } { r + 2 } = \frac { r ^ { 3 } - 7 r - 5 } { ( r + 1 ) ( r + 2 ) }$$ (b) Hence prove, using the method of differences, that $$\sum _ { r = 1 } ^ { n } \frac { r ^ { 3 } - 7 r - 5 } { ( r + 1 ) ( r + 2 ) } = \frac { n \left( n ^ { 2 } + a n + b \right) } { 2 ( n + 2 ) }$$ where \(a\) and \(b\) are constants to be found.

1. (a) Express as a simplified single fraction \(\frac { 1 } { r ^ { 2 } } - \frac { 1 } { ( r + 1 ) ^ { 2 } }\) (b) Hence prove, by the method of differences, that

$$\sum _ { r = 1 } ^ { n } \frac { 2 r + 1 } { r ^ { 2 } ( r + 1 ) ^ { 2 } } = 1 - \frac { 1 } { ( n + 1 ) ^ { 2 } }$$

9

1. Show that \(\frac { 1 } { 3 r - 1 } - \frac { 1 } { 3 r + 2 } \equiv \frac { 3 } { ( 3 r - 1 ) ( 3 r + 2 ) }\).
2. Hence show that \(\sum _ { r = 1 } ^ { 2 n } \frac { 1 } { ( 3 r - 1 ) ( 3 r + 2 ) } = \frac { n } { 2 ( 3 n + 1 ) }\).

2

1. Show that \(\frac { 1 } { r + 1 } - \frac { 1 } { r + 2 } = \frac { 1 } { ( r + 1 ) ( r + 2 ) }\).
2. Hence use the method of differences to find the sum of the series $$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( r + 1 ) ( r + 2 ) }$$

7

1. Show that \(\frac { 1 } { 3 r - 1 } - \frac { 1 } { 3 r + 2 } \equiv \frac { 3 } { ( 3 r - 1 ) ( 3 r + 2 ) }\) for all integers \(r\).
2. Hence use the method of differences to find \(\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 3 r - 1 ) ( 3 r + 2 ) }\). Section B (36 marks)

5

1. Show that \(\frac { 1 } { 5 r - 2 } - \frac { 1 } { 5 r + 3 } \equiv \frac { 5 } { ( 5 r - 2 ) ( 5 r + 3 ) }\) for all integers \(r\).
2. Hence use the method of differences to show that \(\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 5 r - 2 ) ( 5 r + 3 ) } = \frac { n } { 3 ( 5 n + 3 ) }\).

5 Use the result \(\frac { 1 } { 5 r - 1 } - \frac { 1 } { 5 r + 4 } \equiv \frac { 5 } { ( 5 r - 1 ) ( 5 r + 4 ) }\) and the method of differences to find $$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 5 r - 1 ) ( 5 r + 4 ) }$$ simplifying your answer.

5 Given that \(\frac { 3 } { ( 3 r - 1 ) ( 3 r + 2 ) } \equiv \frac { 1 } { 3 r - 1 } - \frac { 1 } { 3 r + 2 }\), find \(\sum _ { r = 1 } ^ { 20 } \frac { 1 } { ( 3 r - 1 ) ( 3 r + 2 ) }\), giving your answer as an exact fraction.

5

1. Show that \(\frac { 1 } { 2 r + 1 } - \frac { 1 } { 2 r + 3 } \equiv \frac { 2 } { ( 2 r + 1 ) ( 2 r + 3 ) }\).
2. Use the method of differences to find \(\sum _ { r = 1 } ^ { 30 } \frac { 1 } { ( 2 r + 1 ) ( 2 r + 3 ) }\), expressing your answer as a fraction.

5 You are given that \(\frac { 4 } { ( 4 n - 3 ) ( 4 n + 1 ) } \equiv \frac { 1 } { 4 n - 3 } - \frac { 1 } { 4 n + 1 }\). Use the method of differences to show that $$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 4 r - 3 ) ( 4 r + 1 ) } = \frac { n } { 4 n + 1 }$$

4 Use the identity \(\frac { 1 } { 2 r + 3 } - \frac { 1 } { 2 r + 5 } \equiv \frac { 2 } { ( 2 r + 3 ) ( 2 r + 5 ) }\) and the method of differences to find \(\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r + 3 ) ( 2 r + 5 ) }\), expressing your answer as a single fraction.

15 (b) Use the method of differences to show that $$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( r + 2 ) ( r + 3 ) } = \frac { n } { 3 ( n + 3 ) }$$ Find the exact solution to the equation $$\sinh \theta ( \sinh \theta + \cosh \theta ) = 1$$

7

1. Show that, for all integers \(r\), $$\frac { 1 } { 2 r - 1 } - \frac { 1 } { 2 r + 1 } = \frac { 2 } { ( 2 r - 1 ) ( 2 r + 1 ) }$$ 7
2. Hence, using the method of differences, show that $$\sum _ { r = 1 } ^ { n } \frac { 1 } { ( 2 r - 1 ) ( 2 r + 1 ) } = \frac { a n } { b n + c }$$ where \(a\), \(b\) and \(c\) are integers to be determined.
   7
3. Hence, or otherwise, evaluate $$\frac { 1 } { 1 \times 3 } + \frac { 1 } { 3 \times 5 } + \frac { 1 } { 5 \times 7 } + \ldots + \frac { 1 } { 99 \times 101 }$$

3
0
2 \end{array} \right] \quad \left[ \begin{array} { c } 5
- 1
3 \end{array} \right] \quad \left[ \begin{array} { l } 2
1
1 \end{array} \right]$$ 2 Use the definitions of \(\cosh x\) and \(\sinh x\) in terms of \(\mathrm { e } ^ { x }\) and \(\mathrm { e } ^ { - x }\) to show that \(\cosh ^ { 2 } x - \sinh ^ { 2 } x \equiv 1\)\\[0pt] [2 marks]\\ 3

1. Given that $$\frac { 2 } { ( r + 1 ) ( r + 2 ) ( r + 3 ) } \equiv \frac { A } { ( r + 1 ) ( r + 2 ) } + \frac { B } { ( r + 2 ) ( r + 3 ) }$$ find the values of the integers \(A\) and \(B\)\\ 3
2. Use the method of differences to show clearly that $$\sum _ { r = 9 } ^ { 97 } \frac { 1 } { ( r + 1 ) ( r + 2 ) ( r + 3 ) } = \frac { 89 } { 19800 }$$

1

1. Use the method of differences to show that \(\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } \left( \frac { 1 } { \mathrm { r } } - \frac { 1 } { \mathrm { r } + 1 } \right) = 1 - \frac { 1 } { \mathrm { n } + 1 }\).
2. Hence determine the following sums.
   1. \(\quad \sum _ { r = 1 } ^ { 99 } \frac { 1 } { r } - \frac { 1 } { r + 1 }\)
   2. \(\quad \sum _ { r = 100 } ^ { \infty } \frac { 1 } { r } - \frac { 1 } { r + 1 }\)

5. (a) Show that \(\frac { 2 } { n - 1 } - \frac { 2 } { n + 1 }\) can be expressed as \(\frac { 4 } { \left( n ^ { 2 } - 1 \right) }\).
(b) Hence, find an expression for \(\sum _ { r = 2 } ^ { n } \frac { 4 } { \left( r ^ { 2 } - 1 \right) }\) in the form \(\frac { ( a n + b ) ( n + c ) } { n ( n + 1 ) }\), where \(a , b , c\) are integers whose values are to be determined.
(c) Explain why \(\sum _ { r = 1 } ^ { 100 } \frac { 4 } { \left( r ^ { 2 } - 1 \right) }\) cannot be calculated.