floor function sum

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Vivek @ Born this Way ·2011-11-16 20:58:23

2009?

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1708

man111 singh ·2011-11-17 03:06:41

Vivek Solution plz.

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sri 3 ·2011-11-17 08:33:52

Since all these terms are between 1 and 2, each term can be taken as 1+{x} where {x} is the frac part.
So √2 = 1+ {√2 } etc.
So A = 2009 + {√2 } + {³√3/2 } + .....
So [A] = 2009 + [ {√2 } + {³√3/2 } + ..... + ]
I dunno wat to do next..

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Vivek @ Born this Way ·2011-11-18 01:13:37

I just thought that after this (after sri's step) the sum in the box function on the right always <1.

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sri 3 ·2011-11-18 04:36:48

Well even I assumed that sum to be < 1. Each term in the sum is lesser than the preceding term but I still dont understand why the sum has to be < 1.

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Basic Integral Calculus Operators Symbols Matrix Cases

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

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∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω

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floor function sum

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