等差数列an=2n-1,bn=(-1)∧(n-1)×4n/anan+1,求bn前n项和.
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等差数列an=2n-1,bn=(-1)∧(n-1)×4n/anan+1,求bn前n项和.
n = (-1)^(n-1) .4n/[an.a(n+1)]
= (-1)^(n-1) .4n/[(2n-1)(2n+1)]
=(-1)^(n-1) .[1/(2n-1)+ 1/(2n+1)]
Tn =b1+b2+b3+...+bn
= ( 1/1 +1/3) -(1/3+1/5)+(1/5+1/7)-(1/7+1/9) +.+(-1)^(n-1) .[1/(2n-1)+ 1/(2n+1)]
= 1 +[(-1)^(n-1)/(2n+1)]
= (-1)^(n-1) .4n/[(2n-1)(2n+1)]
=(-1)^(n-1) .[1/(2n-1)+ 1/(2n+1)]
Tn =b1+b2+b3+...+bn
= ( 1/1 +1/3) -(1/3+1/5)+(1/5+1/7)-(1/7+1/9) +.+(-1)^(n-1) .[1/(2n-1)+ 1/(2n+1)]
= 1 +[(-1)^(n-1)/(2n+1)]
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