Question

数学题在线等急

数学题在线等急过程详细谢谢

Answer 1

待续

追答

错位法求和，

追问

第一步看不懂。。

追答

追问

不好意思 再问一下 那个q平方是怎么算出来的(ﾟДﾟ)ﾉ

追答

两式消除

Answer 2

(I)
an = a1.q^(n-1) ; an>0
a3.a9=(4a5)^2
a1^2 .q^10 = 16(a1)^2. q^8
q^2 = 16
q=4
a2=1
a1.q=1
a1= 1/4
an = (1/4) .4^(n-1) = 4^(n-2)
(II)
let
S=1.4^0+2.4^1+...+n.4^(n-1) (1)
4S= 1.4^1+2.4^2+...+n.4^n (2)
(2)-(1)
3S = n.4^n -( 4^0+4^1+...+4^(n-1))
=n.4^n - (1/4)( 4^n -1)
S = (1/3)[n.4^n - (1/4)( 4^n -1)]
bn
=2n. an
= 2[ n . 4^(n-2)]
=(1/2) [ n . 4^(n-1)]
Sn
=b1+b2+...+bn
=(1/2)S
=(1/6)[n.4^n - (1/4)( 4^n -1)]
=(1/24)[4n.4^n - ( 4^n -1)]
=(1/24)[ 1+ (4n-1).4^n ]