Series/binomial/multinomial theorum

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Homework Statement



for any positive interget n determine

a) SUM(from i=0 to n) of

(-1)^i
i!(n-i)!

b) SUM (from i=0 to n) of

1
i!(n-1)!

Homework Equations





The Attempt at a Solution



a) well i didnt realyl know how to start. i found that with
n=1 then it becomes 1+1
n=2 .5+1+.5
n=3 1/3! + .5 +.5 +1/3!
n=4 1/4! +1/3! +.25 +1/3! +1/4!
n=4 1/5! +1/4! +.083333 +.083333 +1/4! +1/5!

so i guess what i need is how do i get the middle term?
or am i going about it wrong?

for b) i know the sum is always 0, because ichecked it on my calculator. but i don't know hwo to prive that
 
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Do you know what the binomial expansion of (1+x)n is?
 
i acctulay think i got it
a) 2^n/n!
 
Yes. Do the binomial expansion of (1+x)n, and the set x=1 or -1 to get the two versions of your problem.
 
so is the 2^n/n! right?
i get the same answer doing that as i do doing it long form (calculator)

(mind helping on my other post ? :P)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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