已知数列{an}满足:an+1=2an+n-1(n∈N*),a1=1;(1)求数列{an}的通项公式an;(2)设bn=nan,求Sn=b1+b2+…+bn.
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已知数列{an}满足:an+1=2an+n-1(n∈N*),a1=1; (1)求数列{an}的通项公式an; (2)设bn=nan,求Sn=b1+b2+…+bn. |
答案
(1)因为an+1=2an+n-1(n∈N*),所以an+1+(n+1)=2(an+n)(n∈N*), 所以数列{an+n}是以a1+1为首项,2为公比的等比数列, 所以an+n=2n,即an=2n-n. (2)bn=nan=n2n-n2,设Cn=n2n,它的前n项和为Tn, 则Tn=1×2+2×22+3×23+…+n×2n,…① 2Tn=1×22+2×23+3×24+…+n×2n+1…② ②-①得,Tn=-2-(22+23+…+2n)+n×2n+1=(n-1)2n+1+2 所以Sn=b1+b2+…+bn=(n-1)2n+1+2-n(n+1) (2n+1). |
举一反三
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