grayfalcon89 Posted January 4, 2005 Share Posted January 4, 2005 Evaluate: [math]\sum^{49}_{k=1} (-1)^{k} \binom {99}{2k}[/math] Figure this out without using calculator at all. Link to comment Share on other sites More sharing options...
psi20 Posted March 20, 2005 Share Posted March 20, 2005 Sorry, tried to make it look more mathematical. Link to comment Share on other sites More sharing options...
uncool Posted March 22, 2005 Share Posted March 22, 2005 The answer should be (98 49) - 98 - 99. If I'm wrong, please pots it. -Uncool- Link to comment Share on other sites More sharing options...
Algebracus Posted March 22, 2005 Share Posted March 22, 2005 Instead of the old problem, I want to first solve the general problem of finding a short-form of [MATH]S(n) = \sum^{n}_{k=0} (-1)^{k} \binom{2n+1}{2k}[/MATH], where [MATH]n[/MATH] is a natural number of any kind. The result of interest is [MATH]S(49) - 1[/MATH]. We start looking at the expression [MATH](1 - i)^m + (1 + i)^m[/MATH]. By the binomial formula, this can be written as [MATH]\sum^{m}_{k = 0} ((-i)^k + i^k) \binom{m}{k}[/MATH]. For all odd [MATH]k[/MATH] we have [MATH](-i)^k + i^k = 0[/MATH]. For all [MATH]k = 4l[/MATH] we have [MATH](-i)^k + i^k = 2[/MATH]. For all [MATH]k = 4l + 2[/MATH] we have [MATH](-i)^k + i^k = -2[/MATH]. Therefore, the sum we look upon can be rewritten as [MATH]2\binom{m}{0} - 2\binom{m}{2} + 2\binom{m}4 + ...[/MATH] Therefore, [MATH]2S(n) = (1 - i)^{2n + 1} + (1 + i)^{2n + 1}[/MATH]. Furthermore, we know that [MATH]1 - i = \sqrt{2} e^{\frac{7\pi}{4} i}[/MATH] and [MATH]1 + i = \sqrt{2} e^{\frac{\pi}{4} i}[/MATH], so [MATH]2S(n) = \sqrt{2}^{2n + 1}(e^{\frac{7(2n + 1)\pi}{4} i} + e^{\frac{(2n + 1)\pi}{4} i})[/MATH] [MATH]=\sqrt{2}^{2n + 1}(e^{\frac{-(2n + 1)\pi}{4} i} + e^{\frac{(2n + 1)\pi}{4} i})[/MATH]. We divide the cases into two; [MATH]n[/MATH] odd and [MATH]n[/MATH] even. In the first case, [MATH]2S(n) = \sqrt{2}^{2n + 1}(\cos \frac{3\pi}{4} + \cos \frac{-3\pi}{4} + i \sin \frac{3\pi}{4} + i \sin \frac{-3\pi}{4})[/MATH] [MATH]= -\sqrt{2}^{2n + 2} = -2^{n+1}[/MATH]. In the latter case, [MATH]2S(n) = \sqrt{2}^{2n + 1}(\cos \frac{\pi}{4} + \cos \frac{-\pi}{4} + i \sin \frac{\pi}{4} + i \sin \frac{-\pi}{4})[/MATH] [MATH]= \sqrt{2}^{2n + 2} = 2^{n+1}[/MATH]. By this we find [MATH]S(49) - 1 = -2^{49} - 1[/MATH]. Link to comment Share on other sites More sharing options...
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