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Consider the sequence defined recursively by $x_1$=$\sqrt2$ and where $x_n$=$\sqrt2$ + $x_n$$_-$$_1$.

Find a explicit formula for the $n^t$$^h$ term.

I considered using the general equation to find an explicit formula for any term in an arithmetic sequence. a$_n$ = a$_1$ + $d(n-1)$, but I came to no conclusion helping my argument.

Am I using the correct method?

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    $\begingroup$ Should it be $x_n = \sqrt{2} + x_{n-1}$? $\endgroup$
    – naslundx
    Feb 26 '14 at 20:47
  • $\begingroup$ yes, sorry.. minor typo! $\endgroup$ Feb 26 '14 at 20:49
  • $\begingroup$ If you already know this is an arithmetic seq. then you have the general formula: $$x_n=x_1+(n-1)d=\sqrt2+(n-1)\sqrt2=n\sqrt2$$ $\endgroup$
    – DonAntonio
    Feb 26 '14 at 20:51
  • $\begingroup$ $d=x_n-x_{n-1}=\sqrt{2}=x_1=a_1$ use in $a_n=a_1+d(n-1)$ $\endgroup$ Feb 26 '14 at 20:52
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You are right in that it is an arithmetic series. A good strategy is to write up the first terms, simplify, and try to find a pattern:

$$a_1 = \sqrt{2}$$ $$a_2 = \sqrt{2} + a_1 = 2\sqrt{2}$$ $$a_3 = \sqrt{2} + a_2 = \sqrt{2} + 2\sqrt{2} = 3\sqrt{2}$$ $$a_4 = \sqrt{2} + a_3 = \sqrt{2} + 3\sqrt{2} = 4\sqrt{2}$$

And so on. Hence we see the pattern:

$$a_n = n \cdot \sqrt{2}$$

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Here is an approach.

$$ x_{n+1}-x_{n}=\sqrt{2} \implies \sum_{i=0}^{n-1}( x_{i+1}-x_{i}) = \sqrt{2}\sum_{i=0}^{n-1}1 $$

$$ \implies x_n-x_0=\sqrt{2} n .$$

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