Let x 1 = 1/2 and x k + 1 = x k 2 + x k . Find ⌊ ∑ k = 1 100 1/ ( x k + 1 ) ⌋ .

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Let $x_{1} = 1/2$ and $x_{k + 1} = x_{k}^{2} + x_{k}$ . Find $⌊ \sum_{k = 1}^{100} 1/ (x_{k} + 1) ⌋$ .

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Answer:

Solution:

$1/ (x_{k} + 1) = 1/ x_{k} - 1/ x_{k + 1}$ , so the sum is $1/ x_{1} - 1/ x_{101} = 2 - 1/ x_{101}$ . Since $x_{2} = 3/4, x_{3} = 21/16 > 1$ and subsequent terms increase, $0 < 1/ x_{101} < 1$ . The floor is 1.

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:Special sequences/series

⏱ 2mℹ️ Source: QB

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