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A sequence $a_1, a_2, a_3, ...$ 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 7 - 2011 - Paper 1

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Question 7

A-sequence-$a_1,-a_2,-a_3,-...$-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 7-2011-Paper 1.png

A sequence $a_1, a_2, a_3, ...$ 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 t... show full transcript

Worked Solution & Example Answer:A sequence $a_1, a_2, a_3, ...$ 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 7 - 2011 - Paper 1

Step 1

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

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Answer

a2=5k+3a_2 = 5k + 3

Step 2

Show that $a_3 = 25k + 18$.

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Answer

To find a3a_3, we use the expression for a2a_2:

a3=5a2+3=5(5k+3)+3.a_3 = 5a_2 + 3 = 5(5k + 3) + 3.
Expanding this, we have:

a3=25k+15+3=25k+18.a_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 each term:

  • a1=ka_1 = k
  • a2=5k+3a_2 = 5k + 3
  • a3=25k+18a_3 = 25k + 18
  • For a4a_4: a4=5a3+3=5(25k+18)+3=125k+90+3=125k+93.a_4 = 5a_3 + 3 = 5(25k + 18) + 3 = 125k + 90 + 3 = 125k + 93.
    Now we can compute the sum:

n=14an=k+(5k+3)+(25k+18)+(125k+93)=(1+5+25+125)k+(3+18+93)=156k+114.\sum_{n=1}^{4} a_n = k + (5k + 3) + (25k + 18) + (125k + 93) = (1 + 5 + 25 + 125)k + (3 + 18 + 93) = 156k + 114.

Step 4

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

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Answer

We have n=14an=156k+114.\sum_{n=1}^{4} a_n = 156k + 114.
Factoring this: 156k+114=6(26k+19).156k + 114 = 6(26k + 19).
Thus, ( \sum_{n=1}^{4} a_n ) is divisible by 6 since it can be expressed as 6 times an integer.

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