Evaluate:

 

[math]\sum^{49}_{k=1} (-1)^{k} \binom {99}{2k}[/math]

 

Figure this out without using calculator at all.

Sorry, tried to make it look more mathematical.

The answer should be (98 49) - 98 - 99.

 

If I'm wrong, please pots it.

-Uncool-

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].