a1=b1=1,a（n+1）=bn+n,b(n+1)=an+(-1)^2,求证a2n=n^2+n 求1/a2+1/a4+

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a1=b1=1,a（n+1）=bn+n,b(n+1)=an+(-1)^2,求证a2n=n^2+n 求1/a2+1/a4+a/a6+…1/a2n的值

1)
a1=1,b1=1
a2=b1+1=2,b2=a1+(-1)^n=0
a3=b2+2=2,b3=a2+1=3
a4=b3+3=6,b4=a3-1=1
a5=b4+4=5,b5=a4+1=7
a6=b5+5=12,b6=a5-1=4
.
a2n=n(n+1)=n^2+n
2)
1/a2+1/a4+1/a6+...+1/a2n
=1/1*2+1/2*3+1/3*4+...+1/n*(n+1)
=1-1/2+1/2-1/3+1/3-1/4+...+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)

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