组合恒等式的证明：C(r,r)+C(r+1,r)+C(r+2,r)+…+C(n,r)=C(n+1,r+1) C(n,1)+2C(n,2)+…+nC(n,n)=n2^(n-1)还有：C(m,r)*C(n,0)+C(m,r-1)*C(n,1)+…+C(m,0)*C(n,r)=C(m+n,r) (C(n,o))^2+(C(n,1))^2+(C(n,2))^2+(C(n,3))^2+…+(C(n,n))^2=C(2n,n)
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组合恒等式的证明：C(r,r)+C(r+1,r)+C(r+2,r)+…+C(n,r)=C(n+1,r+1) C(n,1)+2C(n,2)+…+nC(n,n)=n2^(n-1)还有：C(m,r)*C(n,0)+C(m,r-1)*C(n,1)+…+C(m,0)*C(n,r)=C(m+n,r) (C(n,o))^2+(C(n,1))^2+(C(n,2))^2+(C(n,3))^2+…+(C(n,n))^2=C(2n,n)
组合恒等式的证明：C(r,r)+C(r+1,r)+C(r+2,r)+…+C(n,r)=C(n+1,r+1) C(n,1)+2C(n,2)+…+nC(n,n)=n2^(n-1)
还有：C(m,r)*C(n,0)+C(m,r-1)*C(n,1)+…+C(m,0)*C(n,r)=C(m+n,r)
(C(n,o))^2+(C(n,1))^2+(C(n,2))^2+(C(n,3))^2+…+(C(n,n))^2=C(2n,n)
组合恒等式的证明：C(r,r)+C(r+1,r)+C(r+2,r)+…+C(n,r)=C(n+1,r+1) C(n,1)+2C(n,2)+…+nC(n,n)=n2^(n-1)还有：C(m,r)*C(n,0)+C(m,r-1)*C(n,1)+…+C(m,0)*C(n,r)=C(m+n,r) (C(n,o))^2+(C(n,1))^2+(C(n,2))^2+(C(n,3))^2+…+(C(n,n))^2=C(2n,n)
1.C(r,r)+C(r+1,r)+C(r+2,r)+…+C(n,r)=C(r+1,r+1)+C(r+1,r)+C(r+2,r)+.+C(n,r)
=C(r+2,r+1)+C(r+2,r)+...+C(n,r)=C(r+3,r+1)+.+C(n,r)=C(n+1,r+1)
2.C(n,1)+2C(n,2)+…+nC(n,n)=nC(n-1,0)+nC(n-1,1)+.+nC(n-1,n-1)
=n[C(n-1,0)+C(n-1,1)+...C(n-1,n-1)]=n*2^(n-1)
3.∵(1+x)^m*(1+x)^n=(1+x)^(m+n)
∴展开式中x^r的系数,右边=C(m+n,r) 左边=C(m,r)*C(n,0)+C(m,r-1)*C(n,1)+…+C(m,0)*C(n,r)
得证
4.(C(n,o))^2+(C(n,1))^2+(C(n,2))^2+(C(n,3))^2+…+(C(n,n))^2=C(n,0)*C(n,n)+C(n,1)*C(n,n-1)+.+C(n,n)*C(n,0)
∵(1+x)^n*(1+x)^n=(1+x)^2n
∴展开式中x^n的系数,右边=C(2n,n)
左边=C(n,0)*C(n,n)+C(n,1)*C(n,n-1)+.+C(n,n)*C(n,0)
得证
注：,看起来有点难,特别是3,4题,但不难理解,
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