Prove recurrence relation formula

6 A sequence is defined by \(u _ { 1 } = 8\) and \(u _ { n + 1 } = 3 u _ { n } + 2 n + 5\). Prove by induction that \(u _ { n } = 4 \left( 3 ^ { n } \right) - n - 3\).

2 For the sequence \(u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots\), it is given that \(u _ { 1 } = 1\) and \(u _ { r + 1 } = \frac { 3 u _ { r } - 2 } { 4 }\) for all \(r\). Prove by mathematical induction that \(u _ { n } = 4 \left( \frac { 3 } { 4 } \right) ^ { n } - 2\), for all positive integers \(n\).

9 For the sequence \(u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots\), it is given that \(u _ { 1 } = 8\) and $$u _ { r + 1 } = \frac { 5 u _ { r } - 3 } { 4 }$$ for all \(r\).

1. Prove by mathematical induction that $$u _ { n } = 4 \left( \frac { 5 } { 4 } \right) ^ { n } + 3$$ for all positive integers \(n\).
2. Deduce the set of values of \(x\) for which the infinite series $$\left( u _ { 1 } - 3 \right) x + \left( u _ { 2 } - 3 \right) x ^ { 2 } + \ldots + \left( u _ { r } - 3 \right) x ^ { r } + \ldots$$ is convergent.
3. Use the result given in part (i) to find surds \(a\) and \(b\) such that $$\sum _ { n = 1 } ^ { N } \ln \left( u _ { n } - 3 \right) = N ^ { 2 } \ln a + N \ln b .$$

3 The irrational number \(\phi = \frac { 1 } { 2 } ( 1 + \sqrt { 5 } )\) plays a significant role in the sequence of Fibonacci numbers given by \(\mathrm { F } _ { 0 } = 0 , \mathrm {~F} _ { 1 } = 1\) and \(\mathrm { F } _ { \mathrm { n } + 1 } = \mathrm { F } _ { \mathrm { n } } + \mathrm { F } _ { \mathrm { n } - 1 }\) for \(n \geqslant 1\). Prove by induction that, for each positive integer \(n , \phi ^ { n } = \mathrm { F } _ { \mathrm { n } } \times \phi + \mathrm { F } _ { \mathrm { n } - 1 }\).

7 The sequence \(u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots\) is defined by $$u _ { 1 } = 2 , \quad u _ { k + 1 } = 2 u _ { k } + 1$$

1. Prove by induction that, for all \(n \geqslant 1\), $$u _ { n } = 3 \times 2 ^ { n - 1 } - 1$$
2. Show that $$\sum _ { r = 1 } ^ { n } u _ { r } = u _ { n + 1 } - ( n + 2 )$$

4 The sequence \(u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots\) is defined by $$u _ { 1 } = \frac { 3 } { 4 } \quad u _ { n + 1 } = \frac { 3 } { 4 - u _ { n } }$$ Prove by induction that, for all \(n \geqslant 1\), $$u _ { n } = \frac { 3 ^ { n + 1 } - 3 } { 3 ^ { n + 1 } - 1 }$$

3 The sequence \(u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots\) is defined by $$u _ { 1 } = 2 , \quad u _ { n + 1 } = \frac { 5 u _ { n } - 3 } { 3 u _ { n } - 1 }$$ Prove by induction that, for all integers \(n \geqslant 1\), $$u _ { n } = \frac { 3 n + 1 } { 3 n - 1 }$$ (6 marks)

5 You are given that \(u _ { 1 } = 5\) and \(u _ { n + 1 } = u _ { n } + 2 n + 4\).
Prove by induction that \(u _ { n } = n ^ { 2 } + 3 n + 1\) for all positive integers \(n\).

3 A sequence is defined by \(a _ { 1 } = 6\) and \(a _ { n + 1 } = 5 a _ { n } - 2\) for \(n \geqslant 1\).
Prove by induction that for all integers \(n \geqslant 1 , a _ { n } = \frac { 11 \times 5 ^ { n - 1 } + 1 } { 2 }\).

A series of positive integers \(u_1, u_2, u_3, \ldots\) is defined by $$u_1 = 6 \text{ and } u_{n+1} = 6u_n - 5, \text{ for } n \geq 1.$$ Prove by induction that \(u_n = 5 \times 6^{n-1} + 1\), for \(n \geq 1\). [5]

The sequence \(u_1, u_2, u_3, \ldots\) is defined by $$u_1 = 1$$ $$u_{n+1} = u_n + 3n$$ Prove by induction that for all integers \(n \geq 1\) $$u_n = \frac{1}{2}n^2 - \frac{3}{2}n + 1$$ [4 marks]

The sequence \(u_1, u_2, u_3, \ldots\) is defined by $$u_1 = 0 \quad u_{n+1} = \frac{5}{6 - u_n}$$ Prove by induction that, for all integers \(n \geq 1\), $$u_n = \frac{5^n - 5}{5^n - 1}$$ [6 marks]

A sequence \(u_n\) is defined by \(u_{n+1} = 2u_n + 3\) and \(u_1 = 1\). Prove by induction that \(u_n = 4 \times 2^{n-1} - 3\) for all positive integers \(n\). [5]

The sequence \(u_1, u_2, u_3, \ldots\) is defined by $$u_1 = 0 \quad u_{n+1} = \frac{5}{6 - u_n}$$ Prove by induction that, for all integers \(n \geq 1\), $$u_n = \frac{5^n - 5}{5^n - 1}$$ [6 marks]