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A sequence $a_n, a_{n+1}, a_{n+2},\, \ldots$ is defined by $a_n = 4a_{n-1} - 3, \quad n > 1$ $a_1 = k,$ where $k$ is a positive integer - Edexcel - A-Level Maths Pure - Question 5 - 2014 - Paper 2

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A-sequence-$a_n,-a_{n+1},-a_{n+2},\,-\ldots$-is-defined-by---$a_n-=-4a_{n-1}---3,-\quad-n->-1$---$a_1-=-k,$-where-$k$-is-a-positive-integer-Edexcel-A-Level Maths Pure-Question 5-2014-Paper 2.png

A sequence $a_n, a_{n+1}, a_{n+2},\, \ldots$ is defined by $a_n = 4a_{n-1} - 3, \quad n > 1$ $a_1 = k,$ where $k$ is a positive integer. (a) Write down an exp... show full transcript

Worked Solution & Example Answer:A sequence $a_n, a_{n+1}, a_{n+2},\, \ldots$ is defined by $a_n = 4a_{n-1} - 3, \quad n > 1$ $a_1 = k,$ where $k$ is a positive integer - Edexcel - A-Level Maths Pure - Question 5 - 2014 - Paper 2

Step 1

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

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Answer

To find the expression for a2a_2, we use the given formula:

a2=4a13a_2 = 4a_1 - 3
Substituting a1=ka_1 = k, we have:

a2=4k3.a_2 = 4k - 3.

Step 2

find the value of $k.$

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Answer

We know that:

n=13an=a1+a2+a3=66.\sum_{n=1}^3 a_n = a_1 + a_2 + a_3 = 66.
Replacing a1a_1 and a2a_2 with their expressions:

a3=4a23=4(4k3)3=16k123=16k15.a_3 = 4a_2 - 3 = 4(4k - 3) - 3 = 16k - 12 - 3 = 16k - 15.
Now substituting these into the summation:

k+(4k3)+(16k15)=66.k + (4k - 3) + (16k - 15) = 66.
Combining like terms:

21k18=66.21k - 18 = 66.
Solving for kk:

21k=84    k=4.21k = 84 \implies k = 4.
Thus, the value of kk is 44.

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