Prove summation formula

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1. Prove by mathematical induction that, for all positive integers \(n\), $$\sum _ { r = 1 } ^ { n } \left( 5 r ^ { 4 } + r ^ { 2 } \right) = \frac { 1 } { 2 } n ^ { 2 } ( n + 1 ) ^ { 2 } ( 2 n + 1 )$$
2. Use the result given in part (a) together with the List of formulae (MF19) to find \(\sum _ { \mathrm { r } = 1 } ^ { \mathrm { n } } \mathrm { r } ^ { 4 }\) in terms of \(n\), fully factorising your answer.

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1. Prove by mathematical induction that, for all positive integers \(n\), $$\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )$$ \includegraphics[max width=\textwidth, alt={}, center]{7eb2abb1-68f4-4cc8-8314-f436906d6c4e-08_2716_35_143_2012} The sum \(S _ { n }\) is defined by \(S _ { n } = \sum _ { r = 1 } ^ { n } r ^ { 4 }\).
2. Using the identity $$( 2 r + 1 ) ^ { 5 } - ( 2 r - 1 ) ^ { 5 } \equiv 160 r ^ { 4 } + 80 r ^ { 2 } + 2$$ show that \(S _ { n } = \frac { 1 } { 30 } n ( n + 1 ) ( 2 n + 1 ) \left( 3 n ^ { 2 } + 3 n - 1 \right)\).
3. Find the value of \(\lim _ { n \rightarrow \infty } \left( n ^ { - 5 } S _ { n } \right)\).

1. (i) Prove by induction that, for \(n \in \mathbb { Z } ^ { + }\)

$$\sum _ { r = 1 } ^ { n } \left( 4 r ^ { 3 } - 3 r ^ { 2 } + r \right) = n ^ { 3 } ( n + 1 )$$ (ii) Prove by induction that, for \(n \in \mathbb { Z } ^ { + }\) $$f ( n ) = 5 ^ { 2 n } + 3 n - 1$$ is divisible by 9

9. (a) Prove by induction that, for \(n \in \mathbb { N }\) $$\sum _ { r = 1 } ^ { n } r ^ { 3 } = \frac { 1 } { 4 } n ^ { 2 } ( n + 1 ) ^ { 2 }$$ (b) Using the standard summation formulae, show that $$\sum _ { r = 1 } ^ { n } r ( r + 1 ) ( r - 1 ) = \frac { 1 } { 4 } n ( n + A ) ( n + B ) ( n + C )$$ where \(A , B\) and \(C\) are constants to be determined.
(c) Determine the value of \(n\) for which $$3 \sum _ { r = 1 } ^ { n } r ( r + 1 ) ( r - 1 ) = 17 \sum _ { r = n } ^ { 2 n } r ^ { 2 }$$

9. (i) Prove by induction that, for \(n \in \mathbb { Z } ^ { + }\) $$\sum _ { r = 1 } ^ { n } r ( r + 1 ) ( r + 2 ) = \frac { n ( n + 1 ) ( n + 2 ) ( n + 3 ) } { 4 }$$ (ii) Prove by induction that, $$4 ^ { n } + 6 n + 8 \text { is divisible by } 18$$ for all positive integers \(n\). \includegraphics[max width=\textwidth, alt={}, center]{df5ab400-5cb1-4b51-8b0a-52dc3587f81a-16_62_44_2476_1889}

1. (i) Prove by induction that, for \(n \in \mathbb { Z } ^ { + }\),

$$\sum _ { r = 1 } ^ { n } r ^ { 2 } ( 2 r - 1 ) = \frac { 1 } { 6 } n ( n + 1 ) \left( 3 n ^ { 2 } + n - 1 \right)$$ (ii) Prove by induction that, for \(n \in \mathbb { Z } ^ { + }\), $$\left( \begin{array} { c c } 7 & - 12 \\ 3 & - 5 \end{array} \right) ^ { n } = \left( \begin{array} { c c } 6 n + 1 & - 12 n \\ 3 n & 1 - 6 n \end{array} \right)$$

8. (a) Prove by induction that, for any positive integer \(n\), $$\sum _ { r = 1 } ^ { n } r ^ { 3 } = \frac { 1 } { 4 } n ^ { 2 } ( n + 1 ) ^ { 2 }$$ (b) Using the formulae for \(\sum _ { r = 1 } ^ { n } r\) and \(\sum _ { r = 1 } ^ { n } r ^ { 3 }\), show that $$\sum _ { r = 1 } ^ { n } \left( r ^ { 3 } + 3 r + 2 \right) = \frac { 1 } { 4 } n ( n + 2 ) \left( n ^ { 2 } + 7 \right)$$ (c) Hence evaluate \(\sum _ { r = 15 } ^ { 25 } \left( r ^ { 3 } + 3 r + 2 \right)\)

6. (a) Prove by induction $$\sum _ { r = 1 } ^ { n } r ^ { 3 } = \frac { 1 } { 4 } n ^ { 2 } ( n + 1 ) ^ { 2 }$$ (b) Using the result in part (a), show that $$\sum _ { r = 1 } ^ { n } \left( r ^ { 3 } - 2 \right) = \frac { 1 } { 4 } n \left( n ^ { 3 } + 2 n ^ { 2 } + n - 8 \right)$$ (c) Calculate the exact value of \(\sum _ { r = 20 } ^ { 50 } \left( r ^ { 3 } - 2 \right)\).

9. (a) Prove by induction that $$\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )$$ Using the standard results for \(\sum _ { r = 1 } ^ { n } r\) and \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\),
(b) show that $$\sum _ { r = 1 } ^ { n } ( r + 2 ) ( r + 3 ) = \frac { 1 } { 3 } n \left( n ^ { 2 } + a n + b \right) ,$$ where \(a\) and \(b\) are integers to be found.
(c) Hence show that $$\sum _ { r = n + 1 } ^ { 2 n } ( r + 2 ) ( r + 3 ) = \frac { 1 } { 3 } n \left( 7 n ^ { 2 } + 27 n + 26 \right)$$

8. (a) Prove by induction, that for \(n \in \mathbb { Z } ^ { + }\), $$\sum _ { r = 1 } ^ { n } r ( 2 r - 1 ) = \frac { 1 } { 6 } n ( n + 1 ) ( 4 n - 1 )$$ (b) Hence, show that $$\sum _ { r = n + 1 } ^ { 3 n } r ( 2 r - 1 ) = \frac { 1 } { 3 } n \left( a n ^ { 2 } + b n + c \right)$$ where \(a\), \(b\) and \(c\) are integers to be found.

2 Prove by induction that, for \(n \geqslant 1 , \sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )\).

2 Prove by induction that, for \(n \geqslant 1 , \sum _ { r = 1 } ^ { n } r ^ { 3 } = \frac { 1 } { 4 } n ^ { 2 } ( n + 1 ) ^ { 2 }\).

6 Prove by induction that \(\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )\).

6 Prove by induction that \(\sum _ { r = 1 } ^ { n } r ^ { 3 } = \frac { 1 } { 4 } n ^ { 2 } ( n + 1 ) ^ { 2 }\).

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1. Using the standard formulae for \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\) and \(\sum _ { r = 1 } ^ { n } r ^ { 3 }\), prove that $$\sum _ { r = 1 } ^ { n } r ^ { 2 } ( r + 1 ) = \frac { 1 } { 12 } n ( n + 1 ) ( n + 2 ) ( 3 n + 1 )$$
2. Prove the same result by mathematical induction.

7. (a) Prove by induction that for \(n \in \mathbb { N }\) $$\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { n } { 6 } ( n + 1 ) ( 2 n + 1 )$$ (b) Hence show that $$\sum _ { r = 1 } ^ { n } \left( r ^ { 2 } + 2 \right) = \frac { n } { 6 } \left( a n ^ { 2 } + b n + c \right)$$ where \(a , b\) and \(c\) are integers to be found.
(c) Using your answers to part (b), find the value of $$\sum _ { r = 10 } ^ { 25 } \left( r ^ { 2 } + 2 \right)$$

1 Prove by induction that, for \(n \geqslant 1 , \sum _ { r = 1 } ^ { n } r ( r + 1 ) = \frac { 1 } { 3 } n ( n + 1 ) ( n + 2 )\).

4 Prove by induction that, for \(n \geqslant 1 , \sum _ { r = 1 } ^ { n } r ( 3 r + 1 ) = n ( n + 1 ) ^ { 2 }\).

6 Prove by induction that \(1 \times 2 + 2 \times 3 + \ldots + n ( n + 1 ) = \frac { n ( n + 1 ) ( n + 2 ) } { 3 }\) for all positive integers \(n\). Section B (36 marks)

6 Prove by induction that \(1 ^ { 2 } - 2 ^ { 2 } + 3 ^ { 2 } - 4 ^ { 2 } + \ldots + ( - 1 ) ^ { n - 1 } n ^ { 2 } = ( - 1 ) ^ { n - 1 } \frac { n ( n + 1 ) } { 2 }\).

6 Prove by induction that \(3 + 10 + 17 + \ldots + ( 7 n - 4 ) = \frac { 1 } { 2 } n ( 7 n - 1 )\) for all positive integers \(n\). Section B (36 marks)

6 Prove by induction that \(1 + 8 + 27 + \ldots + n ^ { 3 } = \frac { 1 } { 4 } n ^ { 2 } ( n + 1 ) ^ { 2 }\). Section B (36 marks)

7 Prove by induction that $$\sum _ { r = 1 } ^ { n } \left( 3 r ^ { 5 } + r ^ { 3 } \right) = \frac { 1 } { 2 } n ^ { 3 } ( n + 1 ) ^ { 3 }$$ for all \(n \geqslant 1\). Use this result together with the List of Formulae (MF10) to prove that $$\sum _ { r = 1 } ^ { n } r ^ { 5 } = \frac { 1 } { 12 } n ^ { 2 } ( n + 1 ) ^ { 2 } \mathrm { Q } ( n )$$ where \(\mathrm { Q } ( n )\) is a quadratic function of \(n\) which is to be determined.

Prove by induction that $$\sum _ { n = 1 } ^ { N } n ^ { 3 } = \frac { 1 } { 4 } N ^ { 2 } ( N + 1 ) ^ { 2 }$$ Use this result, together with the formula for \(\sum _ { n = 1 } ^ { N } n ^ { 2 }\), to show that $$\sum _ { n = 1 } ^ { N } \left( 20 n ^ { 3 } + 36 n ^ { 2 } \right) = N ( N + 1 ) ( N + 3 ) ( 5 N + 2 ) .$$ Let $$S _ { N } = \sum _ { n = 1 } ^ { N } \left( 20 n ^ { 3 } + 36 n ^ { 2 } + \mu n \right)$$ Find the value of the constant \(\mu\) such that \(S _ { N }\) is of the form \(N ^ { 2 } ( N + 1 ) ( a N + b )\), where the constants \(a\) and \(b\) are to be determined. Show that, for this value of \(\mu\), $$5 + \frac { 22 } { N } < N ^ { - 4 } S _ { N } < 5 + \frac { 23 } { N }$$ for all \(N \geqslant 18\).

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1. Prove by induction that, for all integers \(n \geq 1\), $$\sum _ { r = 1 } ^ { n } r ^ { 2 } = \frac { 1 } { 6 } n ( n + 1 ) ( 2 n + 1 )$$ [4 marks]
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2. Hence, or otherwise, write down a factorised expression for the sum of the first \(2 n\) squares $$1 ^ { 2 } + 2 ^ { 2 } + 3 ^ { 2 } + \ldots + ( 2 n ) ^ { 2 }$$ 13
3. Use the formula in part (a) to write down a factorised expression for the sum of the first \(n\) even squares $$2 ^ { 2 } + 4 ^ { 2 } + 6 ^ { 2 } + \ldots + ( 2 n ) ^ { 2 }$$ 13
4. Hence, or otherwise, show that the sum of the first \(n\) odd squares is $$a n ( b n - 1 ) ( b n + 1 )$$ where \(a\) and \(b\) are rational numbers to be determined.