Re: [其他] sum的問題
※ 引述《Danielhsieh (阿光)》之銘言：
: sum x=0 to y x/[(x-1)!(y-x)!]
: 查過答案是(y+1)*2^(y-1)/(y-1)!
: 中間過程想不出來
: 要怎麼把原式的x消掉?
as x/[(x-1)!(y-x)!] = 1/(y-1)! * x * C(y-1, x-1)
so sum_{x=0 to y} { 1/(y-1)! * x * C(y-1, x-1) }
= 1/(y-1)! * sum_{x=0 to y} {x * C(y-1, x-1)}
= 1/(y-1)! * sum_{x=1 to y} {x * C(y-1, x-1)}
= 1/(y-1)! * sum_{x=0 to y-1} {(x+1) * C(y-1,x)}
= 1/(y-1)! * { sum_{x=0 to y-1}C(y-1,x) + sum_{x=0 to y-1}{x*C(y-1,x)} }
= 1/(y-1)! * {2^(y-1) + sum_{x=0 to y-1} x* C(y-1,x) }
now let
f(z) = (1+z)^(y-1) = sum_{x=0 to y-1} C(y-1,x)*z^x
f'(z) = (y-1)*(1+z)^(y-2) = sum_{x=0 to y-1} x*C(y-1,x)*z^(x-1)
set z = 1, so
sum_{x=0 to y-1} {x C(y-1,x)} = (y-1)*2^(y-2)
the answer is (y+1)*2^(y-2)/(y-1)!
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04/09 22:11, , 1F
04/09 22:11, 1F
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