设数列{an}，{bn}（bn>0,n∈N+),满足an=(lgb1+lgb2+...+lgbn)/n (n∈N+),证明:{an}为等差数列的充要条件是{bn}为等比数列.
热心网友
an=(lgb1+lgb2+...+lgbn)/n=lg[b1b2..bn]^(1/n)1.{an}为等差数列an=(n-1)d+a1==lgbn=nan-(n-1)a(n-1)=(2d)n+(a1-2d)==bn=10^(a1-2d)*(100^d)^n=={bn}为等比数列.2.{bn}为等比数列.bk=b1*q^(k-1)dlgbk=lgb1+(n-1)lgq==an=lgb1+(n-1)/2lgq=lgb1+(n-1)lg√q=={an}为等差数列.
热心网友
设数列{an}，{bn}（bn0,n∈N+),满足an=(lgb1+lgb2+。。。+lgbn)/n (n∈N+),证明:{an}为等差数列的充要条件是{bn}为等比数列。 1。充分条件：设{bn}为等比数列，其各项间的比值为k，由于(bn0,n∈N+),则有k0。则an=(lgb1+lgb2+……+lgbn)/n=[lgb1+lg(k*b1)+lg(k^2*b1)+……+lg(k^(n-1)*b1)]/n=[nlgb1+(1+2+3……+n-1)lgk]/n=lgb1+[(1+2+3……+n-1)/n]*lgk=lgb1+(n-1)/2*lgk=lgb1-lgk/2+n/2*lgka(n+1)-an= [lgb1-lgk/2+(n+1)/2*lgk]-(lgb1-lgk/2+n/2*lgk)=lgk/2=constant。由此可见，当{bn}为等比数列时，{an}必定是等差数列。2。必要条件：{an}为等差数列的必要条件是：Δn=a(n+1)-an=常量。Δn=a(n+1)-an=(lgb1+lgb2+……+lgbn+lgb(n+1))/(n+1)-(lgb1+lgb2+……+lgbn)/n=lg(b1b2……b(n+1))/(n+1)-lg(b1b2……bn)/n=lg{ [(b1b2……b(n+1))^(1/(n+1))]/ [(b1b2……bn)^(1/n)] } n=1时Δ1=lg[√(b1b2) /b1] =lg[√(b2/b1)], 令k=b2/b1，b1=b2/kn=2时Δ2=lg[(b1b2b3)^(1/3)/√(kb1b1)] =lg[(kb1b1b3)^(1/3)/√(kb1b1)] =lg[(kb1b1b3)^(1/3)/ (b1√k)]由Δ1=Δ2可得√k=(kb1b1b3)^(1/3)/ (b1√k)，kb1=(kb1b1b3)^(1/3),(kb1)^3=kb1b1b3,==b3=k^2*b1=kb2,== b3/b2=k同理可导出：Δn=Δ1= lg√k=常量。即：{an}成为等差数列的必要条件是{bn}为等比级数。。