组合数求解:C(0,2014)+C(2,2014)+C(4,2014).+C(2014,2014)=?

爱在ee宫 1年前 已收到1个回答 举报

风云98 幼苗

共回答了20个问题采纳率:90% 举报

C(0,2014)+C(2,2014)+C(4,2014).+C(2014,2014)
=(1+1)^2014
=2^2014

1年前 追问

5

爱在ee宫 举报

那2^0*C(0,2014)+2^2*C(2,2014)+2^4*C(4,2014)......+2^2014*C(2014,2014)=?

举报 风云98

2^0*C(0,2014)+2^【1】*C(2,2014)+2^【2】*C(4,2014)......+2^2014*C(2014,2014) =(2+1)^2014 =3^2014

爱在ee宫 举报

可是我学的是2^0*C(0,2014)+2^1*C(1,2014)+2^2*C(2,2014)+2^3*C(3,2014)......+2^2014*C(2014,2014)=(2+1)^2014=3^2014 而题只给偶数了,这该怎么办? 谢谢.

举报 风云98

2^0*C(0,2014)+2^2*C(2,2014)+2^4*C(4,2014)......+2^2014*C(2014,2014) =2^0*C(0,2014)+2^2*C(2,2014)+2^4*C(4,2014)......+2^2014*C(2014,2014) +2^1*C(1,2014)+2^3*C(3,2014)+。。。+2^2013*C(2013,2014) -2^1*C(1,2014)+2^3*C(3,2014)+。。。+2^2013*C(2013,2014) =(2+1)^2014-2^1*C(1,2014)-2^3*C(3,2014)+。。。-2^2013*C(2013,2014) (2-1)^2014=2^0*C(0,2014)+2^2*C(2,2014)+2^4*C(4,014)......+2^2014*C(2014,2014) -2^1*C(1,2014)-2^3*C(3,2014)+。。。-2^2013*C(2013,2014)=1 则 2^0*C(0,2014)+2^2*C(2,2014)+2^4*C(4,2014)......+2^2014*C(2014,2014) =1+2^1*C(1,2014)+2^3*C(3,2014)+。。。+2^2013*C(2013,2014) A=1+B A-B=1 A+B=(2+1)^2014 A=(3^2014+1)/2
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