已知数列{an}满足2a1+2^2a2+2^3a3+...+2^nan=(2n-1)·2^(n+1) +2❶求a1及其通项公式❷求证:1/(a1)²+1/(a2)²+···+1/(an)²<1/4
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已知数列{an}满足2a1+2^2a2+2^3a3+...+2^nan=(2n-1)·2^(n+1) +2
❶求a1及其通项公式
❷求证:1/(a1)²+1/(a2)²+···+1/(an)²<1/4
(1)
2^1.a1+2^2.a2+2^3.a3+...+2^n.an=(2n-1)·2^(n+1) +2 (1)
n=1
2a1=2^2 +2
a1=3
2^1.a1+2^2.a2+2^3.a3+...+2^(n-1).a(n-1)=(2n-3)·2^n +2 (2)
(1)-(2)
2^n.an=(2n-1)·2^(n+1) -(2n-3)·2^n
an = 2(2n-1) -(2n-3)
= 2n+1
(2)
1/(an)^2 = 1/(2n+1)^2
n=1,1/(a1)^2 =1/9
n=2,1/(a2)^2 =1/25
n=3,1/(a3)^2 =1/49
for n>=4
1/(an)^2 = 1/(2n+1)^2
< 1/[(2n-1)(2n+1)]
= (1/2)[ 1/(2n-1) -1/(2n+1) ]
1/(a1)^2+1/(a2)^2+...+1/(an)^2
=1/9+1/25+1/49+[1/(a4)^2+1/(a5)^2+...+1/(an)^2]
< 1/9 +1/25+ 1/49+(1/2)[ 1/7 - 1/(2n+1) ]
< 1/9 +1/25+ 1/49+1/14
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