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Recall that ex=1+x1!+x22!+x33!+⋯ . e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots. ex=1+1!x+2!x2+3!x3+⋯. Then what is the value of 23!+45!+67!+⋯21!+43!+65!+⋯ ?\large \frac{\frac{2}{3!} + \frac{4}{5!} + \frac{6}{7!} + \cdots}{\frac{2}{1!} + \frac{4}{3!} + \frac{6}{5!} + \cdots}\, ? 1!2+3!4+5!6+⋯3!2+5!4+7!6+⋯?
Notation: !!! is the factorial notation. For example, 8!=1×2×3×⋯×88! = 1\times2\times3\times\cdots\times8 8!=1×2×3×⋯×8.
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{a+b+c=503a+b−c=70\large \begin{cases} a+b+c=50 \\ 3a+b-c=70 \end{cases} ⎩⎨⎧a+b+c=503a+b−c=70
a,ba,ba,b and ccc are positive numbers satisfying the system of equations above.
If the range of 5a+4b+2c 5a + 4b + 2c 5a+4b+2c is (m,n) (m,n) (m,n), what is m+n m+ n m+n?
2016x+2016−x=3\large 2016^{x}+2016^{-x}=32016x+2016−x=3
20166x−2016−6x2016x−2016−x= ?\large \sqrt{\frac{2016^{6x}-2016^{-6x}}{2016^{x}-2016^{-x}}} = \, ?2016x−2016−x20166x−2016−6x=?
S=9−139+1381−136561+1343046721−.......S = \displaystyle\sqrt{9 - \sqrt{\dfrac{13}{9} + \sqrt{\dfrac{13}{81} - \sqrt{\dfrac{13}{6561} + \sqrt{\dfrac{13}{43046721} - .......}}}}}S=9−913+8113−656113+4304672113−.......
If S=abS = \sqrt{\dfrac{a}{b}}S=ba where a,ba, ba,b are both primes, find a+ba + ba+b.
∑j=22016∑k=1j−1kj=12+13+23+14+24+34+⋯+20132016+20142016+20152016= ?\sum_{j=2}^{2016} \sum_{k=1}^{j-1} \dfrac kj = \frac{1}{2}+\frac{1}{3}+\frac{2}{3}+\frac{1}{4}+\frac{2}{4}+\frac{3}{4}+\cdots+\frac{2013}{2016}+\frac{2014}{2016}+\frac{2015}{2016}= \, ? j=2∑2016k=1∑j−1jk=21+31+32+41+42+43+⋯+20162013+20162014+20162015=?
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