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A sequence $a_1, a_2, a_3, \ldots$ is defined by $a_1 = 4$ $a_{n+1} = k(a_n + 2)$ for $n > 1$ where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 6 - 2013 - Paper 1

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A-sequence-$a_1,-a_2,-a_3,-\ldots$-is-defined-by---$a_1-=-4$---$a_{n+1}-=-k(a_n-+-2)$-for-$n->-1$---where-$k$-is-a-constant-Edexcel-A-Level Maths Pure-Question 6-2013-Paper 1.png

A sequence $a_1, a_2, a_3, \ldots$ is defined by $a_1 = 4$ $a_{n+1} = k(a_n + 2)$ for $n > 1$ where $k$ is a constant. (a) Find 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 $a_1 = 4$ $a_{n+1} = k(a_n + 2)$ for $n > 1$ where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 6 - 2013 - Paper 1

Step 1

Find an expression for $a_2$ in terms of $k$.

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Answer

Starting from the given recurrence relation:
an+1=k(an+2)a_{n+1} = k(a_n + 2), we can find a2a_2 as follows:

  1. Substitute n=1n = 1:
    a2=k(a1+2)a_2 = k(a_1 + 2)
  2. Since a1=4a_1 = 4:
    a2=k(4+2)=k(6)a_2 = k(4 + 2) = k(6)
    Therefore, the expression for a2a_2 in terms of kk is:
    a2=6ka_2 = 6k.

Step 2

find the two possible values of $k$.

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Answer

To find the two possible values of kk, we first need to calculate a3a_3:

  1. Using the expression found for a2a_2:
    a3=k(a2+2)=k(6k+2)=6k2+2ka_3 = k(a_2 + 2) = k(6k + 2) = 6k^2 + 2k
  2. Then, calculate the sum i=13ai\sum_{i=1}^{3} a_i:
    a1+a2+a3=4+6k+(6k2+2k)=6k2+8k+4a_1 + a_2 + a_3 = 4 + 6k + (6k^2 + 2k) = 6k^2 + 8k + 4
  3. Set this sum equal to 2:
    6k2+8k+4=26k^2 + 8k + 4 = 2
  4. Rearranging gives:
    6k2+8k+2=06k^2 + 8k + 2 = 0
  5. Simplifying further:
    3k2+4k+1=03k^2 + 4k + 1 = 0
  6. Using the quadratic formula k=b±b24ac2ak = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} with a=3a = 3, b=4b = 4, c=1c = 1:
    k=4±4243123=4±16126=4±26k = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} = \frac{-4 \pm \sqrt{16 - 12}}{6} = \frac{-4 \pm 2}{6}
  7. This leads to two possible values:
    k=26=13andk=66=1.k = \frac{-2}{6} = -\frac{1}{3} \quad \text{and} \quad k = \frac{-6}{6} = -1.
    Thus, the two possible values of kk are 13-\frac{1}{3} and 1-1.

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