Live in past
x n + 1 = 1 + x n + x n 2 x n + x n 2 ; x 1 = 2 1 If { x n } is a sequence of numbers satisfying the recurrence relation above, find x 1 + 1 1 + x 2 + 1 1 + x 3 + 1 1 + . . . . . . x 2 0 1 2 + 1 1 + x 2 0 1 3 1
The answer is 2014.
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1 solution
I Think In the first line of solution you interchanged 1/xn + 1 and 1/xn+1 that is n+1th term and 1 plus the nth term
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No , i just reciprocaled the term and formed a telescoping series (not interchanged) also it would make no diff if i interchange them.
by the way how is the problem's name given(why live in the past??)
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Haha it is coz the series is till 2013 and the answer is 2014.That's why "Live in past" Pls reshare it and upvote solutions u like.I have seen you do not do that .It takes only a click.
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ohh! i will upvote and reshare from now.thanks for pointing out!
x n + 1 1 = 1 + x n 1 − x n + 1 1
Now starting from n=1 till n=2012 and adding these eq, terms starts cancelling and we are left with:
x 2 0 1 3 1 = 2 0 1 2 + x 1 1 − n = 1 ∑ 2 0 1 2 x n + 1 1
Putting x 1 = 2 1
We get x 1 + 1 1 + x 2 + 1 1 + x 3 + 1 1 + . . . . . . x 2 0 1 2 + 1 1 + x 2 0 1 3 1 = 2 0 1 4