A sequence $a_1, a_2, a_3,\ldots$ is defined by a_1 = k, a_{n+1} = 3a_n + 5, where $k$ is a positive integer. (a) Write down an expression for $a_2$ in terms of $k$. (b) Show that $a_3 = 9k + 20$. (c) (i) Find $\sum_{r=1}^4 a_r$ in terms of $k$. (ii) Show that $\sum_{r=1}^4 a_r$ is divisible by 10.

A-Level Edexcel Maths Pure Graphs of Functions: 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} = 3a_n + 5, where $k$ is a positive integer - Edexcel - A-Level Maths Pure - Question 9 - 2007 - Paper 1

Question 9

$A-sequence-$a_1,-a_2,-a_3,\ldots$-is-defined-by----a_1-=-k,----a_{n+1}-=-3a_n-+-5,--where-$k$-is-a-positive-integer-Edexcel-A-Level Maths Pure-Question 9-2007-Paper 1.png$

A sequence $a_1, a_2, a_3,\ldots$ is defined by a_1 = k, a_{n+1} = 3a_n + 5, where $k$ is a positive integer. (a) Write down an expression for $a_2$ in ter... 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} = 3a_n + 5, where $k$ is a positive integer - Edexcel - A-Level Maths Pure - Question 9 - 2007 - 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 recursive formula given:

$a_{n+1} = 3a_n + 5$

Substituting $n=1$ into the formula:

$a_2 = 3a_1 + 5 = 3k + 5.$

Step 2

Show that $a_3 = 9k + 20$

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Answer

Using the expression for $a_2$ obtained previously:

$a_3 = 3a_2 + 5 = 3(3k + 5) + 5 = 9k + 15 + 5 = 9k + 20.$

Step 3

Find $\sum_{r=1}^4 a_r$ in terms of $k$

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Answer

We will calculate each term up to $a_4$ using the recursive relations:

$a_1 = k$
$a_2 = 3k + 5$ (from part (a))
$a_3 = 9k + 20$ (from part (b))
Now to find $a_4$ : $a_4 = 3a_3 + 5 = 3(9k + 20) + 5 = 27k + 60 + 5 = 27k + 65.$

Thus, we have:

$\sum_{r=1}^4 a_r = a_1 + a_2 + a_3 + a_4 = k + (3k + 5) + (9k + 20) + (27k + 65)$
This simplifies to:

$\sum_{r=1}^4 a_r = (k + 3k + 9k + 27k) + (5 + 20 + 65) = 40k + 90.$

Step 4

Show that $\sum_{r=1}^4 a_r$ is divisible by 10

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Answer

To show divisibility by 10, we take the expression:

$\sum_{r=1}^4 a_r = 40k + 90.$
Notice that both terms (40k and 90) are divisible by 10:

$40k$ is divisible by 10 since 40 is already a multiple of 10.
$90$ is also divisible by 10.

Therefore, their sum $40k + 90$ is divisible by 10, confirming that:

$\sum_{r=1}^4 a_r \text{ is divisible by } 10.$

A sequence $a_1, a_2, a_3,\ldots$ is defined by a_1 = k, a_{n+1} = 3a_n + 5, where $k$ is a positive integer - Edexcel - A-Level Maths Pure - Question 9 - 2007 - Paper 1

Worked Solution & Example Answer:A sequence $a_1, a_2, a_3,\ldots$ is defined by a_1 = k, a_{n+1} = 3a_n + 5, where $k$ is a positive integer - Edexcel - A-Level Maths Pure - Question 9 - 2007 - Paper 1

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

Show that $a_3 = 9k + 20$

Find $\sum_{r=1}^4 a_r$ in terms of $k$

Show that $\sum_{r=1}^4 a_r$ is divisible by 10

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