在数列{an}中a1=1,前n项和为Sn=n^2an求sn
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Sn=n^2an ==> (n^2-1)an = S(n-1) = (n-1)^2a(n-1)
==> an = ((n-1)/(n+1))*a(n-1)
= (n-1)/(n+1) * (n-2)/n * a(n-2)
= ....
= (n-1)/(n+1) * (n-2)/n * ...*1/3 * a1
= 2/((n+1)*n) = 2/n - 2/(n+1)
Sn = an + a(n-1) + ... + a1
= 2/n - 2/(n+1) + 2/(n-1) - 2/(n) + ... + 2/1 - 2/2
= 2 - 2/(n+1) = 2n/(n+1)
==> an = ((n-1)/(n+1))*a(n-1)
= (n-1)/(n+1) * (n-2)/n * a(n-2)
= ....
= (n-1)/(n+1) * (n-2)/n * ...*1/3 * a1
= 2/((n+1)*n) = 2/n - 2/(n+1)
Sn = an + a(n-1) + ... + a1
= 2/n - 2/(n+1) + 2/(n-1) - 2/(n) + ... + 2/1 - 2/2
= 2 - 2/(n+1) = 2n/(n+1)
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