已知数列{an}满足a1=1 a2=3 且a(n+2)=(1+2|cosnπ/2|)an+|sinnπ/2|,n为正整数(1)求a(2k-1) 免求(2)数列{yn}{bn}满足yn=a(2n-1),b1=y1 且当n大于等于2时bn=yn^2(1/y1^2+1/y2^2+1/y3^2+…+1/y(n-1)^2) 证明当n
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已知数列{an}满足a1=1 a2=3 且a(n+2)=(1+2|cosnπ/2|)an+|sinnπ/2|,n为正整数(1)求a(2k-1) 免求(2)数列{yn}{bn}满足yn=a(2n-1),b1=y1 且当n大于等于2时bn=yn^2(1/y1^2+1/y2^2+1/y3^2+…+1/y(n-1)^2) 证明当n
已知数列{an}满足a1=1 a2=3 且a(n+2)=(1+2|cosnπ/2|)an+|sinnπ/2|,n为正整数
(1)求a(2k-1) 免求
(2)数列{yn}{bn}满足yn=a(2n-1),b1=y1 且当n大于等于2时
bn=yn^2(1/y1^2+1/y2^2+1/y3^2+…+1/y(n-1)^2)
证明当n大于等于2时,b(n+1)/(n+1)^2-bn/n^2=1/n^2
(3) 在(2)的条件下,试比较(1+1/b1)(1+1/b2)(1+1/b3)……(1+1/bn)与4的大小关系
已知数列{an}满足a1=1 a2=3 且a(n+2)=(1+2|cosnπ/2|)an+|sinnπ/2|,n为正整数(1)求a(2k-1) 免求(2)数列{yn}{bn}满足yn=a(2n-1),b1=y1 且当n大于等于2时bn=yn^2(1/y1^2+1/y2^2+1/y3^2+…+1/y(n-1)^2) 证明当n
(1)
n=2k-1 (取奇数时),|cosnπ/2|= 0 ,|sinnπ/2|=1
a(n+2)=an+1
即n=2k-1时,{an}为等差数列
可求出a(2k-1)
n=2k (取偶数时 ),|cosnπ/2|= 1 ,|sinnπ/2|=0
a(n+2)=3an
即 n=2k时,{an}为等比数列
(2)
{yn}=a(2n-1),{yn}为等差数列,公差为1,y1=a1=1
故通项公式yn=1+(n-1)=n
bn=n²(1/1²+1/2²+…+1/(n-1)²)
bn/n²=(1/1²+1/2²+…+1/(n-1)²)
b(n+1)/(n+1)²=(1/1²+1/2²+…+1/(n-1)²+1/n²)
b(n+1)/(n+1)²-bn/n²
=1/n²
(3)
b(n+1)/(n+1)²-bn/n²=1/n²
n²b(n+1)-(n+1)²bn=(n+1)²
1+bn=b(n+1)n²/(n+1)²
1+1/bn=(1+bn)/bn=b(n+1)n²/bn(n+1)²
(1+1/b1)(1+1/b2)(1+1/b3)……(1+1/b(n-1))(1+1/bn)
=[b2*1²/b1*2²]……[bn(n-1)²/b(n-1)n²][b(n+1)n²/bn(n+1)²]
=[b2*b3*b4*……b(n+1)*1²*2²*3²*4²*……n²]/[b1*b2*……bn*2²*3²*……(n+1)²]
=b(n+1)/[b1*(n+1)²]
=b(n+1)/(n+1)²
由bn=n²(1/1²+1/2²+…+1/(n-1)²)
得
b(n+1)/(n+1)²
=[(n+1)²(1/1²+1/2²+…+1/n²)]/(n+1)²
=1/1²+1/2²+…+1/n²
自然数平方倒数和为π²/6
所以比4小.