作业帮数列的一题求解,谢谢
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数列的一题求解,谢谢
数列的一题求解,谢谢
数列的一题求解,谢谢
设公差为d,则d>0,数列单调递增
a1=a2-d=2
d=a2-2
a3=a2^2 -10
a2+d=a2^2-10
d=a2-2代入,整理,得
a2^2-2a2-8=0
(a2+2)(a2-4)=0
a2=-2(数列为递增数列,a2>a1,舍去)或a2=4
d=a2-2=4-2=2
an=a1+(n-1)d=2+2(n-1)=2n
数列{an}的通项公式为an=2n
y=4[sin(πx+ 1/2)]^2 -1=2[1-cos(2πx +1)] -1=1 -2cos(2πx +1)
最小正周期T=2π/(2π)=1
b1=1
bn=b1q^(n-1)=1×3^(n-1)=3^(n-1)
an-bn=2n -3^(n-1)
Sn=2(1+2+...+n) -[1+3+...+3^(n-1)]
=2n(n+1)/2 -1×(3ⁿ-1)/(3-1)
=n^2 +n -(1/2)×3ⁿ +1/2
f(n)=2/(2n+a1)+2/(2n+a2)+...+2/(2n+an)
f(n+1)=2/[2(n+1)+a1]+2/[2(n+1)+a2]+...+2/[2(n+1)+an]+2/[2(n+1)+a(n+1)]
=2/[2n+(a1+2)]+2/[2n+(a2+2)]+...+2/[2n+(an+2)]+2/[2n+a(n+1)+2]
=2/(2n+a2)+2/(2n+a3)+...+2/[2n+a(n+1)]+2/[2n+a(n+2)]
f(n+1)-f(n)=2/(2n+a2)+2/(2n+a3)+...+2/[2n+a(n+2)]-[2/(2n+a1)+2/(2n+a2)+...+2/(2n+an)]
=2/[2n+a(n+1)]+2/[2n+a(n+2)]-2/(2n+a1)
=2/[2n+2(n+1)]+2/[2n+2(n+2)]-2/(2n+2)
=1/(2n+1)+1/(2n+2)-1/(n+1)
=1/(2n+1) -1/(2n+2)>0
即随n增大,f(n)单调递增
当n=2时,f(n)有最小值
[f(n)]min=f(2)=2/(2×2+a1)+2/(2×2+a2)=2/(4+2)+2/(4+4)=7/12