(1)證明：∵在數(shù)列{a[n]}中,已知a[n]+a[n+1]=2n (n∈N*)
∴用待定系數(shù)法,有：
a[n+1]+x(n+1)+y=-(a[n]+xn+y)
∵-2x=2,-x-2y=0
∴x=-1,y=1/2
∴a[n+1]-(n+1)+1/2=-(a[n]-n+1/2)
∴{a[n]-n+1/2}是首項為a[1]-1/2,公比為-1的等比數(shù)列
即：a[n]-n+1/2=(a[1]-1/2)(-1)^(n-1)
a[n]=(1/2-a[1])(-1)^n+n-1/2
∵2a[2n+1]=-1+2a[1]+4n+2-1=2(2n+a[1])
a[2n-1]=-1/2+a[1]+2n-1-1/2=2n+a[1]-2
a[2n+3]=-1/2+a[1]+2n+3-1/2=2n+a[1]+2
∴2a[2n+1]=a[2n-1]+a[2n+3]
公差=(a[2n+3]-a[2n-1])/2=2
∵2a[2n]=1-2a[1]+4n-1=2(2n-a[1])
a[2n-2]=1/2-a[1]+2n-2-1/2=2n-a[1]-2
a[2n+2]=1/2-a[1]+2n+2-1/2=2n-a[1]+2
∴2a[2n]=a[2n-2]+a[2n+2]
公差=(a[2n+2]-a[2n-2])/2=2
綜上所述：數(shù)列{a[2n+1]},{a[2n]}均成等差數(shù)列,并且公差均為2
(2)∵在數(shù)列{b[n]}中,已知b[n]b[n+1]=2^n (n∈N*)
∴類比上述結(jié)論,有：{b[2n+1]},{b[2n]}均成等比數(shù)列,并求公比
證明：∵在數(shù)列{b[n]}中,已知b[n]b[n+1]=2^n (n∈N*)
∴l(xiāng)nb[n]+lnb[n+1]=nln2
用待定系數(shù)法,有：
lnb[n+1]+x(n+1)+y=-(lnb[n]+xn+y)
∵-2x=ln2,-x-2y=0
∴x=-ln2/2,y=ln2/4
∴l(xiāng)nb[n+1]-(n+1)ln2/2+ln2/4=-(lnb[n]-nln2/2+ln2/4)
∴{lnb[n]-nln2/2+ln2/4}是首項為lnb[1]-ln2/4,公比為-1的等比數(shù)列
即：lnb[n]-nln2/2+ln2/4=(lnb[1]-ln2/4)(-1)^(n-1)
∴b[n]=e^{(ln2/4-lnb[1])(-1)^n+nln2/2-ln2/4}
∵b[2n+1]^2=e^{2(lnb[1]-ln2/4+nln2+ln2/2-ln2/4)}=e^{2(nln2+lnb[1])}
b[2n-1]=e^(lnb[1]-ln2/4+nln2-ln2/2-ln2/4)=e^(nln2+lnb[1]-ln2)
b[2n+3]=e^(lnb[1]-ln2/4+nln2+3ln2/2-ln2/4)=e^(nln2+lnb[1]+ln2)
∴b[2n+1]^2=b[2n-1]b[2n+3]
公比=√(b[2n+3]/b[2n-1])=2
∵b[2n]^2=e^{2(ln2/4-lnb[1]+nln2-ln2/4)}=e^{2(nln2-lnb[1])}
b[2n-2]=e^(ln2/4-lnb[1]+nln2-ln2-ln2/4)=e^(nln2-lnb[1]-ln2)
b[2n+2]=e^(ln2/4-lnb[1]+nln2+ln2-ln2/4)=e^(nln2-lnb[1]+ln2)
∴b[2n]^2=b[2n-2]b[2n+2]
公比=√(b[2n+2]/b[2n-2])=2
綜上所述：數(shù)列{b[2n+1]},{b[2n]}均成等比數(shù)列,并且公比均為2