已知数列an中,a1=1,a2=2,且an+2=an+(-1)^n+1+n,则a2n-a2n-1=?
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a(n+2)=an+(-1)^n +1+n
a(n+2) -an = (-1)^n +1+n
if n is odd number
a(n+2) -an = n
an-a(n-2)=(n-2)
an - a1 = 1+3+5+...+(n-2)
an=1+3+5+..+(n-2) +1
= (n-1)^2/4 +1
if n is even number
a(n+2) -an = n+2
an-a(n-2)=n
an - a2 =4+6+8+..+n
an= 2+4+6+..+n
= (n+2)n/4
a2n-a(2n-1)
= (n+1)n - (n-1)^2-1
= 3n-2
a(n+2) -an = (-1)^n +1+n
if n is odd number
a(n+2) -an = n
an-a(n-2)=(n-2)
an - a1 = 1+3+5+...+(n-2)
an=1+3+5+..+(n-2) +1
= (n-1)^2/4 +1
if n is even number
a(n+2) -an = n+2
an-a(n-2)=n
an - a2 =4+6+8+..+n
an= 2+4+6+..+n
= (n+2)n/4
a2n-a(2n-1)
= (n+1)n - (n-1)^2-1
= 3n-2
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