Proving standard summation formulae

3

1. By considering \(( 2 r + 1 ) ^ { 3 } - ( 2 r - 1 ) ^ { 3 }\), use the method of differences to prove that $$\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )$$ Let \(S _ { n } = 1 ^ { 2 } + 3 \times 2 ^ { 2 } + 3 ^ { 2 } + 3 \times 4 ^ { 2 } + 5 ^ { 2 } + 3 \times 6 ^ { 2 } + \ldots + \left( 2 + ( - 1 ) ^ { n } \right) n ^ { 2 }\).
2. Show that \(\mathrm { S } _ { 2 \mathrm { n } } = \frac { 1 } { 3 } \mathrm { n } ( 2 \mathrm { n } + 1 ) ( \mathrm { an } + \mathrm { b } )\), where \(a\) and \(b\) are integers to be determined.
3. State the value of \(\lim _ { n \rightarrow \infty } \frac { S _ { 2 n } } { n ^ { 3 } }\).

1

1. By considering \(( r + 1 ) ^ { 2 } - r ^ { 2 }\), use the method of differences to prove that $$\sum _ { r = 1 } ^ { n } r = \frac { 1 } { 2 } n ( n + 1 )$$
2. Given that \(\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } ( \mathrm { r } + \mathrm { a } ) = \mathrm { n }\), find \(a\) in terms of \(n\).

3. (a) Show that \(r ^ { 3 } - ( r - 1 ) ^ { 3 } \equiv 3 r ^ { 2 } - 3 r + 1\) (b) Hence prove by the method of differences that, for \(n \in \mathbb { Z } ^ { + }\) $$\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { n ( n + 1 ) ( 2 n + 1 ) } { 6 }$$ [You may use \(\sum _ { r = 1 } ^ { n } r = \frac { n ( n + 1 ) } { 2 }\) without proof.]

2. Given that for all real values of \(r , \quad ( 2 r + 1 ) ^ { 3 } - ( 2 r - 1 ) ^ { 3 } = A r ^ { 2 } + B\), where \(A\) and \(B\) are constants,

1. find the value of \(A\) and the value of \(B\).
2. Hence, or otherwise, prove that \(\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )\).
3. Calculate \(\sum _ { r = 1 } ^ { 40 } ( 3 r - 1 ) ^ { 2 }\).
   (3)(Total 10 marks)

4. Given that $$( 2 r + 1 ) ^ { 3 } = A r ^ { 3 } + B r ^ { 2 } + C r + 1 ,$$

1. find the values of the constants \(A , B\) and \(C\).
2. Show that $$( 2 r + 1 ) ^ { 3 } - ( 2 r - 1 ) ^ { 3 } = 24 r ^ { 2 } + 2$$
3. Using the result in part (b) and the method of differences, show that $$\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )$$

4. (a) Show that $$r ^ { 2 } ( r + 1 ) ^ { 2 } - ( r - 1 ) ^ { 2 } r ^ { 2 } \equiv 4 r ^ { 3 }$$ Given that \(\sum _ { r = 1 } ^ { n } r = \frac { 1 } { 2 } n ( n + 1 )\) (b) use the identity in (a) and the method of differences to show that $$\left( 1 ^ { 3 } + 2 ^ { 3 } + 3 ^ { 3 } + \ldots + n ^ { 3 } \right) = ( 1 + 2 + 3 + \ldots + n ) ^ { 2 }$$

1. (a) Show that

$$n ^ { 5 } - ( n - 1 ) ^ { 5 } \equiv 5 n ^ { 4 } - 10 n ^ { 3 } + 10 n ^ { 2 } - 5 n + 1$$ (b) Hence, using the method of differences, show that for all integer values of \(n\), $$\sum _ { r = 1 } ^ { n } r ^ { 4 } = \frac { 1 } { 30 } n ( n + 1 ) ( 2 n + 1 ) \left( a n ^ { 2 } + b n + c \right)$$ where \(a\), \(b\) and \(c\) are integers to be determined.

9

1. Use the method of differences to show that $$\sum _ { r = 1 } ^ { n } \left\{ ( r + 1 ) ^ { 3 } - r ^ { 3 } \right\} = ( n + 1 ) ^ { 3 } - 1$$
2. Show that \(( r + 1 ) ^ { 3 } - r ^ { 3 } \equiv 3 r ^ { 2 } + 3 r + 1\).
3. Use the results in parts (i) and (ii) and the standard result for \(\sum _ { r = 1 } ^ { n } r\) to show that $$3 \sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 2 } n ( n + 1 ) ( 2 n + 1 )$$

9

1. Show that \(r ( r + 1 ) ( r + 2 ) - ( r - 1 ) r ( r + 1 ) \equiv 3 r ( r + 1 )\).
2. Hence use the method of differences to find an expression for \(\sum _ { r = 1 } ^ { n } r ( r + 1 )\).
3. Show that you can obtain the same expression for \(\sum _ { r = 1 } ^ { n } r ( r + 1 )\) using the standard formulae for \(\sum _ { r = 1 } ^ { n } r\) and \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\).

7

1. Use the method of differences to show that $$\sum _ { r = 1 } ^ { n } \left\{ ( r + 1 ) ^ { 4 } - r ^ { 4 } \right\} = ( n + 1 ) ^ { 4 } - 1$$
2. Show that \(( r + 1 ) ^ { 4 } - r ^ { 4 } \equiv 4 r ^ { 3 } + 6 r ^ { 2 } + 4 r + 1\).
3. Hence show that $$4 \sum _ { r = 1 } ^ { n } r ^ { 3 } = n ^ { 2 } ( n + 1 ) ^ { 2 }$$

2 Expand and simplify \(( r + 1 ) ^ { 4 } - r ^ { 4 }\). Use the method of differences together with the standard results for \(\sum _ { r = 1 } ^ { n } r\) and \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\) to show that $$\sum _ { r = 1 } ^ { n } r ^ { 3 } = \frac { 1 } { 4 } n ^ { 2 } ( n + 1 ) ^ { 2 }$$

4 Let \(\mathrm { f } ( r ) = r ( r + 1 ) ( r + 2 )\). Show that $$\mathrm { f } ( r ) - \mathrm { f } ( r - 1 ) = 3 r ( r + 1 )$$ Hence show that \(\sum _ { r = 1 } ^ { n } r ( r + 1 ) = \frac { 1 } { 3 } n ( n + 1 ) ( n + 2 )\). Using the standard result for \(\sum _ { r = 1 } ^ { n } r\), deduce that \(\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )\). Find the sum of the series $$1 ^ { 2 } + 2 \times 2 ^ { 2 } + 3 ^ { 2 } + 2 \times 4 ^ { 2 } + 5 ^ { 2 } + 2 \times 6 ^ { 2 } + \ldots + 2 ( n - 1 ) ^ { 2 } + n ^ { 2 }$$ where \(n\) is odd.

1

1. By considering \(( r + 1 ) ^ { 3 } - r ^ { 3 }\), find \(\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } \left( 3 \mathrm { r } ^ { 2 } + 3 \mathrm { r } + 1 \right)\).
2. Use this result to find \(\sum _ { r = 1 } ^ { n } r ( r + 1 )\), expressing your answer in fully factorised form.

2

1. Show that $$( 2 r + 1 ) ^ { 3 } - ( 2 r - 1 ) ^ { 3 } = 24 r ^ { 2 } + 2$$
2. Hence, using the method of differences, show that $$\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )$$

5

1. Show that $$r ^ { 2 } ( r + 1 ) ^ { 2 } - ( r - 1 ) ^ { 2 } r ^ { 2 } = p r ^ { 3 }$$ where \(p\) is an integer to be found.
   [0pt] [1 mark]
   5
2. Hence use the method of differences to show that $$\sum _ { r = 1 } ^ { n } r ^ { 3 } = \frac { 1 } { 4 } n ^ { 2 } ( n + 1 ) ^ { 2 }$$