9 You are given that \(\frac { 3 } { 4 ( 2 r - 1 ) } - \frac { 1 } { 2 r + 1 } + \frac { 1 } { 4 ( 2 r + 3 ) } = \frac { 2 r + 5 } { ( 2 r - 1 ) ( 2 r + 1 ) ( 2 r + 3 ) }\).
- Use the method of differences to show that
$$\sum _ { r = 1 } ^ { n } \frac { 2 r + 5 } { ( 2 r - 1 ) ( 2 r + 1 ) ( 2 r + 3 ) } = \frac { 2 } { 3 } - \frac { 3 } { 4 ( 2 n + 1 ) } + \frac { 1 } { 4 ( 2 n + 3 ) } .$$
- Write down the limit to which \(\sum _ { r = 1 } ^ { n } \frac { 2 r + 5 } { ( 2 r - 1 ) ( 2 r + 1 ) ( 2 r + 3 ) }\) converges as \(n\) tends to infinity.
- Find the sum of the finite series
$$\frac { 45 } { 39 \times 41 \times 43 } + \frac { 47 } { 41 \times 43 \times 45 } + \frac { 49 } { 43 \times 45 \times 47 } + \ldots + \frac { 105 } { 99 \times 101 \times 103 } ,$$
giving your answer to 3 significant figures.
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