Question

等比数列{bn}中，b5+b6=b7-b5=48,则S10=_____，请写出过程

Answer 1

设b5=a,公比为q
a+aq=48
aq^2-a=48
两式相除
(1+q)/(q^2-1)=1
1/(q-1)=1
q=2
a=16
又a=b1*q^4
所以b1=1
s10=(1-2^10)/(1-2)
=1023

不懂再问哦

Answer 2

b5+b6=48
b7-b5=48
两式相加得
b6+b7=96
b5q+b6q=96
q(b5+b6)=96
48q=96
q=2
b5+b6=q^4(b1+b2)
48=16(b1+b2)
b1+b2=3
b1+b1q=3
b1(1+q)=3
3b1=3
b1=1

s10=b1(1-q^10)/(1-q)
=(1-2^10)/(1-2)
=2^10-1
=1024-1
=1023