A sequence $x_1, x_2, x_3, \ldots,$ is defined by $x_1 = 1$ $x_{n+1} = (x_n)^2 - kx_n, \ n > 1$ where $k$ is a constant, $k \neq 0$ (a) Find an expression for $x_2$ in terms of $k$. (b) Show that $x_3 = 1 - 3k + 2k^2$ Given also that $x_3 = 1$, (c) calculate the value of $k$. (d) Hence find the value of $\sum_{n=1}^{100} x_n$

A-Level Edexcel Maths Pure Simultaneous Equations: A sequence $x_1, x_2, x

A sequence $x_1, x_2, x_3, \ldots,$ is defined by $x_1 = 1$ $x_{n+1} = (x_n)^2 - kx_n, \ n > 1$ where $k$ is a constant, $k \neq 0$ (a) Find an expression for $x_2$ in terms of $k$ - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 2

Question 7

$A-sequence-$x_1,-x_2,-x_3,-\ldots,$-is-defined-by--$x_1-=-1$---$x_{n+1}-=-(x_n)^2---kx_n,-\-n->-1$---where-$k$-is-a-constant,-$k-\neq-0$----(a)-Find-an-expression-for-$x_2$-in-terms-of-$k$-Edexcel-A-Level Maths Pure-Question 7-2013-Paper 2.png$

A sequence $x_1, x_2, x_3, \ldots,$ is defined by $x_1 = 1$ $x_{n+1} = (x_n)^2 - kx_n, \ n > 1$ where $k$ is a constant, $k \neq 0$ (a) Find an expression fo... show full transcript

Worked Solution & Example Answer:A sequence $x_1, x_2, x_3, \ldots,$ is defined by $x_1 = 1$ $x_{n+1} = (x_n)^2 - kx_n, \ n > 1$ where $k$ is a constant, $k \neq 0$ (a) Find an expression for $x_2$ in terms of $k$ - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 2

Step 1

Find an expression for $x_2$ in terms of $k$

96%

114 rated

Answer

Given $x_1 = 1$ , we can find $x_2$ as follows:

x_2 = (x_1)^2 - kx_1 = 1^2 - k(1) = 1 - k.

Step 2

Show that $x_3 = 1 - 3k + 2k^2$

99%

104 rated

Answer

To find $x_3$ , we use the formula:

x_3 = (x_2)^2 - kx_2.

Substituting $x_2 = 1 - k$ gives:

x_3 = (1 - k)^2 - k(1 - k) = (1 - 2k + k^2) - (k - k^2) = 1 - 2k + k^2 - k + k^2 = 1 - 3k + 2k^2.

Step 3

calculate the value of $k$

96%

101 rated

Answer

Given that $x_3 = 1$ , we can set the equation:

1 - 3k + 2k^2 = 1.

This simplifies to:

-3k + 2k^2 = 0.

Factoring gives:

k(2k - 3) = 0.

Since $k \neq 0$ , we find:

2k - 3 = 0 \Rightarrow k = \frac{3}{2}.

Step 4

Hence find the value of $\sum_{n=1}^{100} x_n$

98%

120 rated

Answer

Using the formula derived earlier, we know:

x_n = (1 - k)^{2^{n-1}} + k \sum_{j=0}^{n-2} 2^{j}

With $k = \frac{3}{2}$ , we find:

1 - k = 1 - \frac{3}{2} = -\frac{1}{2}.

Thus:

S = \sum_{n=1}^{100} x_n = \sum_{n=1}^{100} (1 - k)^{2^{n-1}} + \sum_{n=1}^{100} k.

The sum simplifies to find the total values affecting each term.

A sequence $x_1, x_2, x_3, \ldots,$ is defined by $x_1 = 1$ $x_{n+1} = (x_n)^2 - kx_n, \ n > 1$ where $k$ is a constant, $k \neq 0$ (a) Find an expression for $x_2$ in terms of $k$ - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 2

Worked Solution & Example Answer:A sequence $x_1, x_2, x_3, \ldots,$ is defined by $x_1 = 1$ $x_{n+1} = (x_n)^2 - kx_n, \ n > 1$ where $k$ is a constant, $k \neq 0$ (a) Find an expression for $x_2$ in terms of $k$ - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 2

Find an expression for $x_2$ in terms of $k$

Show that $x_3 = 1 - 3k + 2k^2$

calculate the value of $k$

Hence find the value of $\sum_{n=1}^{100} x_n$

Join the A-Level students using SimpleStudy...