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A sequence $a_1, a_2, a_3,...$ is defined by a_1 = 1 a_{n+1} = \frac{k(a_n + 1)}{a_n}, \, n > 1 where $k$ is a positive constant - Edexcel - A-Level Maths Pure - Question 5 - 2017 - Paper 1

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A-sequence-$a_1,-a_2,-a_3,...$-is-defined-by---a_1-=-1---a_{n+1}-=-\frac{k(a_n-+-1)}{a_n},-\,-n->-1---where-$k$-is-a-positive-constant-Edexcel-A-Level Maths Pure-Question 5-2017-Paper 1.png

A sequence $a_1, a_2, a_3,...$ is defined by a_1 = 1 a_{n+1} = \frac{k(a_n + 1)}{a_n}, \, n > 1 where $k$ is a positive constant. (a) Write down expressions f... show full transcript

Worked Solution & Example Answer:A sequence $a_1, a_2, a_3,...$ is defined by a_1 = 1 a_{n+1} = \frac{k(a_n + 1)}{a_n}, \, n > 1 where $k$ is a positive constant - Edexcel - A-Level Maths Pure - Question 5 - 2017 - Paper 1

Step 1

Write down expressions for $a_2$ and $a_3$ in terms of $k$

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Answer

To find a2a_2, we substitute n=1n = 1 into the formula:

a2=k(a1+1)a1=k(1+1)1=2k.a_2 = \frac{k(a_1 + 1)}{a_1} = \frac{k(1 + 1)}{1} = 2k.

Next, to find a3a_3, we substitute n=2n = 2:

a3=k(a2+1)a2=k(2k+1)2k=k(2k+1)2k=2k+12.a_3 = \frac{k(a_2 + 1)}{a_2} = \frac{k(2k + 1)}{2k} = \frac{k(2k + 1)}{2k} = \frac{2k + 1}{2}.

Step 2

find an exact value for $k$

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Answer

Given that r=13ar=10\sum_{r=1}^3 a_r = 10, we calculate:

a1+a2+a3=1+2k+2k+12=10.a_1 + a_2 + a_3 = 1 + 2k + \frac{2k + 1}{2} = 10.

Combining like terms, we have:

1+2k+2k+12=22+2k+2k+12=2+4k+2k+12=1+6k2=10.1 + 2k + \frac{2k + 1}{2} = \frac{2}{2} + 2k + \frac{2k + 1}{2} = \frac{2 + 4k + 2k + 1}{2} = \frac{1 + 6k}{2} = 10.

Multiplying both sides by 2 gives:

1+6k=206k=19k=196.1 + 6k = 20\Rightarrow 6k = 19\Rightarrow k = \frac{19}{6}.

Thus, the exact value for kk is 196\frac{19}{6}.

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