I = k = 1 ∑ 2 0 1 6 k + 1 + k 1
Find ⌈ I ⌉ .
Notation : ⌈ ⋅ ⌉ denotes the ceiling function .
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Nice and detailed. You've improved on LaTeX writing. Keep up the effort!
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StackExchange and Wikipedia helped. Also, I use a Google Chrome extension called Daum Equation Editor. I like your colourful solutions though, very innovative.
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Thanks. Colorful solutions are fun to write!
celling function . i do the same
Nicely structured!
Relevant wiki: Telescoping Series - Sum
This is a rather simple one. k + 1 + k 1 = ( k + 1 ) 2 − ( k ) 2 k + 1 − k = k + 1 − k Now, I = k = 1 ∑ 2 0 1 6 k + 1 + k 1 = 2 0 1 7 − 2 0 1 6 + 2 0 1 6 − 2 0 1 5 . . . . + 2 − 1 = 2 0 1 7 − 1 This can be calculate to be approximately 4 3 . 9 , which the ceiling function makes 4 4 .
I'm sorry but I dont't know where to ask this: Is the maths written Mathjax?
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It is written in LaTex. Here is a useful LaTex guide. To practice Latex you can use this Google Chrome extension.
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Relevant wiki: Telescoping Series - Sum
I ⌈ I ⌉ = k = 1 ∑ 2 0 1 6 k + 1 + k 1 = k = 1 ∑ 2 0 1 6 k + 1 + k 1 ( k + 1 − k k + 1 − k ) = k = 1 ∑ 2 0 1 6 ( k + 1 + k ) ( k + 1 − k ) 1 ( k + 1 − k ) = k = 1 ∑ 2 0 1 6 ( k + 1 + k ) ( k + 1 − k ) k + 1 − k = k = 1 ∑ 2 0 1 6 ( k + 1 ) 2 − ( k ) 2 k + 1 − k = k = 1 ∑ 2 0 1 6 k + 1 − k k + 1 − k = k = 1 ∑ 2 0 1 6 k + 1 − k = ( 2 0 1 7 − 2 0 1 6 ) + ( 2 0 1 6 − 2 0 1 5 ) + ⋯ + ( 3 − 2 ) + ( 2 − 1 ) = 2 0 1 7 − 2 0 1 6 + 2 0 1 6 − 2 0 1 5 + ⋯ + 3 − 2 + 2 − 1 = 2 0 1 7 − 1 ≈ 4 3 . 9 = ⌈ 4 3 . 9 ⌉ = 4 4 Question Rationalize the denominator Multiply fractions 1 × a = a ( a + b ) ( a − b ) = a 2 − b 2 Simplify and cancel Simplify Remove brackets and cancel Simplify Use ceiling function