已知等比数列{an}的首项a1>0,公比q>0.设数列{bn}的通项bn=a(n+1)+a(n+2),数列{an},{bn}的前n项之和为An和Bn,试比较An和Bn的大小
已知等比数列{an}的首项a1>0,公比q>0.设数列{bn}的通项bn=a(n+1)+a(n+2),数列{an},{bn}的前n项之和为An和Bn,试比较An和Bn的大小
由题意An=a1+a2+a3+……+an ,
Bn=b1+b2+b3+……+ bn= a2+a3+ a3+ a4……+an+ an+1+ an+1+ an+2
=An+ a3+ a4……+an+ an+1+ an+1+ an+2－a1
Bn－An = a3+ a4……+an+ an+1+ an+1+ an+2－a1 = a1 (1－qn )(q2+q-1)/( 1－q)
令 q2+q-1=0 q=(√5－1)／2 ,
∴ 当 0即Bn当q > (√5－1)／2 时,q2+q-1> 0 ,故Bn－An > 0,即Bn> An
为什么Bn－An = a3+ a4……+an+ an+1+ an+1+ an+2－a1 = a1 (1－qn )(q2+q-1)/( 1－q)
(1－qn )(q2+q-1)/( 1－q)哪里来的
a(n) = a(1)q^(n-1).
q不为1时,
s(n) = a(1)[1-q^n]/(1-q).
a(3)+a(4)+...+a(n) + a(n+1) + a(n+1) + a(n+2)- a(1)
= a(3)+a(4)+...+a(n)+a(n+1)+a(n+2) + a(n+1)-a(1).
= a(1)[q^2 + q^3 + ...+ q^(n-1) + q^n + q^(n+1) + q^n - 1]
= a(1)q^2[1 + q + ...+ q^(n-3) + q^(n-2) + q^(n-1)] + a(1)[q^n - 1]
= a(1)q^2[1 - q^n]/(1-q) - a(1)[1 - q^n]
= a(1)q^2 [1-q^n]/(1-q) - a(1)(1-q)[1-q^n]/(1-q)
= a(1)[1-q^n]/(1-q)*[q^2 - 1 + q]
= a(1)[1 - q^n][q^2 + q - 1]/(1-q)