A sequence $a_1, a_2, a_3, \ldots$ is defined by $a_1 = k$, $a_{n+1} = 5a_n + 3$, 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 Integration: A sequence $a_1, a_2, a

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

Worked Solution & Example Answer:A sequence $a_1, a_2, a_3, \ldots$ is defined by $a_1 = k$, $a_{n+1} = 5a_n + 3$, 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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Answer

To find $a_2$ , we use the recurrence relation.

Substituting $n=1$ :
$a_2 = 5a_1 + 3 = 5(k) + 3 = 5k + 3.$

Step 2

Show that $a_3 = 25k + 18$.

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Answer

To find $a_3$ , we will substitute $a_2$ into the recurrence relation.

Substituting $n=2$ :
$a_3 = 5a_2 + 3 = 5(5k + 3) + 3 = 25k + 15 + 3 = 25k + 18.$

Step 3

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

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Answer

We first find $a_4$ .

Using $a_3$ in the recurrence:
$a_4 = 5a_3 + 3 = 5(25k + 18) + 3 = 125k + 90 + 3 = 125k + 93.$

Now we can calculate the sum:
$\sum_{n=1}^{4} a_n = a_1 + a_2 + a_3 + a_4 = k + (5k + 3) + (25k + 18) + (125k + 93).$

Combining the terms,
$\sum_{n=1}^{4} a_n = (1k + 5k + 25k + 125k) + (3 + 18 + 93) = 156k + 114.$

Step 4

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

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Answer

From our previous result, we have:
$\sum_{n=1}^{4} a_n = 156k + 114.$

We can factor this as:
$156k + 114 = 6(26k + 19).$

Thus, it is clear that the sum is divisible by 6.

A sequence $a_1, a_2, a_3, \ldots$ is defined by $a_1 = k$, $a_{n+1} = 5a_n + 3$, 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 $a_1 = k$, $a_{n+1} = 5a_n + 3$, 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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