求证:C0n+3C1n+5C2n+…+(2n+1)Cnn=(n+1)2
来源:学生作业帮 编辑:搜搜考试网作业帮 分类:数学作业 时间:2024/07/03 13:54:46
求证:
+3
+5
+…+(2n+1)
=(n+1)2
C | 0 n |
C | 1 n |
C | 2 n |
C | n n |
![求证:C0n+3C1n+5C2n+…+(2n+1)Cnn=(n+1)2](/uploads/image/z/17577298-10-8.jpg?t=%E6%B1%82%E8%AF%81%EF%BC%9AC0n%2B3C1n%2B5C2n%2B%E2%80%A6%2B%282n%2B1%29Cnn%EF%BC%9D%28n%2B1%292)
证明:设Sn=
C0n+3
C1n+5
C2n+…+(2n+1)
Cnn ①
把①式右边倒转过来得Sn=(2n+1)
Cnn+(2n−1)
Cn−1n+…+3
C1n+
C0n,
又由
Cmn=
Cn−mn可得Sn=(2n+1)
C0n+(2n−1)
C1n+…+3
Cn−1n+
Cnn②
①+②得 2Sn=(2n+2)(
C0n+
C1n+…+
Cn−1n+
Cnn)=2(n+1)•2n,
∴Sn=(n+1)•2n,
即:
C0n+3
C1n+5
C2n+…+(2n+1)
Cnn=(n+1)2n,
原等式得证.
C0n+3
C1n+5
C2n+…+(2n+1)
Cnn ①
把①式右边倒转过来得Sn=(2n+1)
Cnn+(2n−1)
Cn−1n+…+3
C1n+
C0n,
又由
Cmn=
Cn−mn可得Sn=(2n+1)
C0n+(2n−1)
C1n+…+3
Cn−1n+
Cnn②
①+②得 2Sn=(2n+2)(
C0n+
C1n+…+
Cn−1n+
Cnn)=2(n+1)•2n,
∴Sn=(n+1)•2n,
即:
C0n+3
C1n+5
C2n+…+(2n+1)
Cnn=(n+1)2n,
原等式得证.
求证:C0n+2C1n+3C2n+…+(n+1)Cnn
证明一个组合数等式,C0n*3^n+C1n*3^(n-1)+C2n*3^(n-2)+.+Cnn*3^0=(1+3)^n
求证:C0n+2C2n+……+(n+1)Cnn=2∧n+n*2∧(n-1)
排列组合证明题:(C0n)2+ (C1n)2+…+(Cnn)2=(2n!)/n!
求证:(Cn0)*2+(Cn1)*2+…+(Cnn)*2=C2n n
求证Cn0Cn1+Cn1Cn2+……+Cn(n-1)Cnn=(2n)!/(n-1)!(n+1)!
求证:Cn0*Cn1+Cn1*Cn2+.+Cn(n-1)*Cnn=(2n)!/((n-1)!*(n+1)!)
已知三个共点力大小为零,则三个力大小可能是?A .15N.6N.5N B.3N.6N.4N C1N.2N.10 D.1N
数学二项式定力求证:Cn0/1+Cn1/2+Cn2/3……+Cnn/n+1={2^(n+1)-1}/(n+1)
2[Cn0+2Cn1+3Cn2+…+(n+1)Cnn] =(n+2)(Cn0+Cn1+…Cnn)怎么来的
Cn0+2Cn1+3Cn2+…+(n+1)Cnn=256求n的值
求证:Cn1+Cn2+.+Cnn=1+2+2^2+.+2^(n-1)