C0 10-(1/2)*C1 10+(1/3)*C2 10-(1/4)*C3 10+.+(1/11)*C10 10的值为____
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![C0 10-(1/2)*C1 10+(1/3)*C2 10-(1/4)*C3 10+.+(1/11)*C10 10的值为____](/uploads/image/z/8424411-51-1.jpg?t=C0+10-%281%2F2%29%2AC1+10%2B%281%2F3%29%2AC2+10-%281%2F4%29%2AC3+10%2B.%2B%281%2F11%29%2AC10+10%E7%9A%84%E5%80%BC%E4%B8%BA____)
C0 10-(1/2)*C1 10+(1/3)*C2 10-(1/4)*C3 10+.+(1/11)*C10 10的值为____
C0 10-(1/2)*C1 10+(1/3)*C2 10-(1/4)*C3 10+.+(1/11)*C10 10的值为____
C0 10-(1/2)*C1 10+(1/3)*C2 10-(1/4)*C3 10+.+(1/11)*C10 10的值为____
法一
抽取每一项
(-1)^k (1/k+1) c(10,k)=(-1)^k[10!/(k+1)!(10-k)!]= (-1/11)* {(-1)^(k+1) [11!/(k+1)!(10-k)!]}
=(-1/11)[(-1)^(k+1) *C(11,k+1)]
所以原式=(-1/11)[-C(11,1)+C(11,2)+...-C(11,11)]
=(-1/11) {[1+(-1)]^11-1}
=1/11
就是把每一项,比如(1/3)C(10,2)=10!/[3!8!]=(1/11)[11!/(3!8!)]=(1/11)C(11,3)
然后把这个算式,变成了(1-1)^11的二项展开式相关的算式.
如果学过微积分的话,这个
法二
如果学过微积分,这题还有个简单的办法.
原式=∫(0到1) [(1-x)^10] dx (令t=1-x)
=∫((0到1) [t^10] dt
=1/11
C(m,n)=C(m,m-n)C(m+1,n)=C(m,n)+C(m,n-1)C(2,1)+C(3,2)+……+C(10,9)=C(2,1)+C(3,1)+……+C(10,1)=C(2,2)+C(2,1)+C(3,1)+……+C(10,1)-1=C(3,2)+C(3,1)+……+C(10,1)-1……=C(11,2)-1=54