已知等差数列an的前n项和为sn,且a2=3,a5=9,数列bn=sn-n求bn的前n 项和tn 急急急 10
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d=(a5-a2)/(5-2)=2
an=a2+(n-2)d=3+2(n-2)=2n-1
bn=sn-n
=(a1+an)n/2-n
=(1+2n-1)n/2-n
=n²-n
Tn=b1+b2+b3+……+bn
=1²-1+2²-2+3²-3+……+n²-n
=(1²+2²+3²+……+n²)-(1+2+3+……+n)
=n(n+1)(2n+1)/6-(1+n)n/2
=n(n+1)(2n+1-3)/6
=n(n+1)(n-1)/3
=n(n²-1)/3
an=a2+(n-2)d=3+2(n-2)=2n-1
bn=sn-n
=(a1+an)n/2-n
=(1+2n-1)n/2-n
=n²-n
Tn=b1+b2+b3+……+bn
=1²-1+2²-2+3²-3+……+n²-n
=(1²+2²+3²+……+n²)-(1+2+3+……+n)
=n(n+1)(2n+1)/6-(1+n)n/2
=n(n+1)(2n+1-3)/6
=n(n+1)(n-1)/3
=n(n²-1)/3
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