S
=
n
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0
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6
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Find the value of
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1
⌋
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Notation : ⌊ ⋅ ⌋ denotes the floor function .
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I also had -(2017*2017)/2016, However, when I punched that into the calculator, I got 2018.00496. Which would imply that the answer is 2018. This seems a bit strange
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⌊ 2 0 1 8 . 0 0 4 9 6 ⌋ = 2 0 1 8 but ⌊ − 2 0 1 8 . 0 0 4 9 6 ⌋ = − 2 0 1 9 . Note that ⌊ 0 . 5 ⌋ = 0 but ⌊ − 0 . 5 ⌋ = − 1 .
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Thank you man. I was entering the answer with my phone, and it did not give options to include negative answers, so I thought I had messed up the sign somewhere, or that the answer required absolute values.
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@Ceesay Muhammed – You can use Wolfram Alpha on phone. It is free. You should key in floor(-2018.00496), it should give you the right answer. I am using Excel spreadsheet -- =INT(-2018.00486).
Here, S = − 2 0 1 6 1 + 2 0 1 6 2 2 − 2 0 1 6 3 3 + … Now, 2 0 1 6 S = − 2 0 1 6 2 1 + 2 0 1 6 3 2 + … Adding these two equations, we get ( 1 + 2 0 1 6 1 ) S = − 2 0 1 6 1 + 2 0 1 6 2 1 − 2 0 1 6 3 1 − … Using the formalism of geometric progression we have S = − 2 0 1 7 2 2 0 1 6
S = n = 0 ∑ ∞ 2 0 1 6 n ( − 1 ) n n = − ( 2 0 1 6 1 + 2 0 1 6 3 3 + 2 0 1 6 5 5 + ⋯ ) + ( 2 0 1 6 2 2 + 2 0 1 6 4 4 + 2 0 1 6 6 6 + ⋯ ) Clearly, this is the sum of two Arithmetic-Geometric Progressions, which are easily evaluated, giving us the answer -2019.
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n = 0 ∑ ∞ ( − x ) n d x d n = 0 ∑ ∞ ( − x ) n n = 0 ∑ ∞ ( − x ) n − 1 n n = 0 ∑ ∞ ( − x ) n n n = 0 ∑ ∞ 2 0 1 6 n ( − 1 ) n n ⟹ S ⟹ ⌊ S 1 ⌋ = 1 + x 1 Maclaurin series, for − 1 < x < 1 = d x d ( 1 + x 1 ) = − ( 1 + x ) 2 1 Multiplying both sides by x = − ( 1 + x ) 2 x Putting x = 2 0 1 6 1 = − ( 1 + 2 0 1 6 ) 2 2 0 1 6 = − ( 1 + 2 0 1 6 ) 2 2 0 1 6 = ⌊ 2 0 1 6 ( 2 0 1 6 + 1 ) 2 ⌋ = ⌊ 2 0 1 6 2 0 1 6 2 + 2 ( 2 0 1 6 ) + 1 ⌋ = − 2 0 1 9