A sequence $a_1, a_2, a_3, \ldots$ is defined by $a_1 = k$, $a_{n+1} = 5a_n + 3$, $n \geq 1$, where $k$ is a positive integer. (a) Write down an expression for $a_2$ in terms of $k$. (b) Show that $a_3 = 25k + 18$. (c) (i) Find $\sum_{n=1}^{4} a_n$ in terms of $k$, in its simplest form. (ii) Show that $\sum_{n=1}^{4} a_n$ is divisible by 6.

A-Level Edexcel Maths Pure Differentiation: A sequence $a_1, a_2, a

A sequence $a_1, a_2, a_3, \ldots$ is defined by $a_1 = k$, $a_{n+1} = 5a_n + 3$, $n \geq 1$, where $k$ is a positive integer - Edexcel - A-Level Maths Pure - Question 6 - 2011 - Paper 1

Question 6

$A-sequence-$a_1,-a_2,-a_3,-\ldots$-is-defined-by----$a_1-=-k$,---$a_{n+1}-=-5a_n-+-3$,-$n-\geq-1$,---where-$k$-is-a-positive-integer-Edexcel-A-Level Maths Pure-Question 6-2011-Paper 1.png$

A sequence $a_1, a_2, a_3, \ldots$ is defined by $a_1 = k$, $a_{n+1} = 5a_n + 3$, $n \geq 1$, where $k$ is a positive integer. (a) Write down an expression f... show full transcript

Worked Solution & Example Answer:A sequence $a_1, a_2, a_3, \ldots$ is defined by $a_1 = k$, $a_{n+1} = 5a_n + 3$, $n \geq 1$, where $k$ is a positive integer - Edexcel - A-Level Maths Pure - Question 6 - 2011 - Paper 1

Step 1

Write down an expression for $a_2$ in terms of $k$.

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114 rated

Answer

To find $a_2$ , we use the recursive formula:

a_2 = 5a_1 + 3
Since $a_1 = k$ , we get:

a_2 = 5k + 3.

Step 2

Show that $a_3 = 25k + 18$.

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104 rated

Answer

Using the recursive definition again,

a_3 = 5a_2 + 3.
Substituting for $a_2$ gives:

a_3 = 5(5k + 3) + 3 = 25k + 15 + 3 = 25k + 18.

Step 3

Find $\sum_{n=1}^{4} a_n$ in terms of $k$, in its simplest form.

96%

101 rated

Answer

We calculate each term first:

$a_1 = k$
$a_2 = 5k + 3$
$a_3 = 25k + 18$
$a_4 = 5a_3 + 3 = 5(25k + 18) + 3 = 125k + 90 + 3 = 125k + 93$ .

Now sum these:

$\sum_{n=1}^{4} a_n = a_1 + a_2 + a_3 + a_4$
$= k + (5k + 3) + (25k + 18) + (125k + 93)$
$= k + 5k + 25k + 125k + 3 + 18 + 93$
$= 156k + 114.$

Step 4

Show that $\sum_{n=1}^{4} a_n$ is divisible by 6.

98%

120 rated

Answer

We evaluate:
$\sum_{n=1}^{4} a_n = 156k + 114.$
Notice:

156 is divisible by 6.
114 is also divisible by 6. This means:
$\sum_{n=1}^{4} a_n = 6(26k + 19),$
which shows that the entire expression is divisible by 6.

A sequence $a_1, a_2, a_3, \ldots$ is defined by $a_1 = k$, $a_{n+1} = 5a_n + 3$, $n \geq 1$, where $k$ is a positive integer - Edexcel - A-Level Maths Pure - Question 6 - 2011 - Paper 1

Worked Solution & Example Answer:A sequence $a_1, a_2, a_3, \ldots$ is defined by $a_1 = k$, $a_{n+1} = 5a_n + 3$, $n \geq 1$, where $k$ is a positive integer - Edexcel - A-Level Maths Pure - Question 6 - 2011 - Paper 1

Write down an expression for $a_2$ in terms of $k$.

Show that $a_3 = 25k + 18$.

Find $\sum_{n=1}^{4} a_n$ in terms of $k$, in its simplest form.

Show that $\sum_{n=1}^{4} a_n$ is divisible by 6.

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