设{an}是a1=4的单调递增数列,且满足an+1^2+an^2+16=8(an+1+an)+2an+1an,求an
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设{an}是a1=4的单调递增数列,且满足an+1^2+an^2+16=8(an+1+an)+2an+1an,求an
n+1均为a的下标
n+1均为a的下标
这很容易.
因为an+1^2+ an^2+16=8 (an+1 + an)+ 2 an+1 an
所以(an+1 + an )^2 -8(an+1 + an)+16=
4 an+1 an
即(an+1 + an -4)^2=4 an+1 an
因为a1=4,且单增,所以开方得
an+1 + an -4=2(an+1 an)^(1/2)
所以(an+1)^(1/2) -(an)^(1/2)=2
所以数列{(an)^(1/2)}成等差数列
求得(an)^(1/2)=2n
所以an=4n^2
因为an+1^2+ an^2+16=8 (an+1 + an)+ 2 an+1 an
所以(an+1 + an )^2 -8(an+1 + an)+16=
4 an+1 an
即(an+1 + an -4)^2=4 an+1 an
因为a1=4,且单增,所以开方得
an+1 + an -4=2(an+1 an)^(1/2)
所以(an+1)^(1/2) -(an)^(1/2)=2
所以数列{(an)^(1/2)}成等差数列
求得(an)^(1/2)=2n
所以an=4n^2
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