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$$^h$ term.
I considered using the general equation to find an explicit formula for any term in an arithmetic sequence. a$_n$ = a$_1$ + $d(n-1)$, but I came to no conclusion helping my argument.
Am I using the correct method?
$\endgroup$ 42 Answers
$\begingroup$Here is an approach.
$$ 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 .$$
$\endgroup$ $\begingroup$You are right in that it is an arithmetic series. A good strategy is to write up the first terms, simplify, and try to find a pattern:
$$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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