Answer :

LammettHash

Multiply 2 by 1/2 to get 1.

Multiply 1 by 2/3 to get 2/3.

Multiply 2/3 by 3/4 to get 6/12 = 1/2.

Multiply 1/2 by 4/5 to get 4/10 = 2/5.

Multiply 2/5 by 5/6 to get 10/30 = 1/3.

Multiply 1/3 by 6/7 to get 6/21 = 2/7. (I suspect there's a typo in the question.)

And so on, so that the nth term in the sequence is multiplied by n/(n + 1) to get the (n + 1)th term.

Recursively, the sequence is given by

[tex]\begin{cases}a_1=2\\a_n=\dfrac{n-1}na_{n-1}&\text{for }n>1\end{cases}[/tex]

We can solve this exactly by iterating:

[tex]a_n=\dfrac{n-1}na_{n-1}=\dfrac{n-1}n\dfrac{n-2}na_{n-1}=\dfrac{n-1}n\dfrac{n-2}{n-1}\dfrac{n-3}{n-2}a_{n-3}=\cdots[/tex]

and so on down to

[tex]a_n=\dfrac{(n-1)\cdot(n-2)\cdot(n-3)\cdot\cdots\cdot3\cdot2\cdot1}{n\cdot(n-1)\cdot(n-2)\cdot\cdots\cdot4\cdot3\cdot2}a_1[/tex]

or

[tex]a_n=\dfrac{(n-1)!}{n!}a_1[/tex]

and with lots of cancellation, we end up with

[tex]a_n=\dfrac{a_1}n=\boxed{\dfrac2n}[/tex]

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