已知数列{an}中,a1=1,an+1=2an+1,令bn=an+1-an.
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已知数列{an}中,a1=1,an+1=2an+1,令bn=an+1-an.
(1)证明:数列{bn}是等比数列;
(2)设数列{nan}的前n项和为Sn,求使Sn+
(1)证明:数列{bn}是等比数列;
(2)设数列{nan}的前n项和为Sn,求使Sn+
n(n+1) |
2 |
(1)证明:由an+1=2an+1,得an=2an-1+1(n≥2),
两式相减得:(an+1-an)=2(an-an-1).
∵bn=an+1-an,
∴bn=2bn-1.
又b1=a2-a1=(2a1+1)-a1=a1+1=2.
∴数列{bn}是以2为首项,以2为公比等比数列;
(2)由(1)得bn=2n,即an+1−an=2n,
∴an=a1+(a2−a1)+(a3−a2)+…+(a n−an−1)=1+2+22+…2n−1=2n−1,
∴nan=n•2n−n,
∴Sn=(1•21−1)+(2•22−1)+…+(n•2n−n)=(1•21+2•22+…+n•2n)−
n(n+1)
2,
令T=1•21+2•22+…+n•2n ①,
则2T=1•22+2•23+…+(n-1)•2n+n•2n+1 ②,
①-②得:-T=-2+2n+1-n•2n+1,
∴T=(n-1)•2n+1+2,
∴Sn=(n−1)•2n+1+2−
n(n+1)
2,
由Sn+
n(n+1)
2>120,得(n-1)•2n+1+2>120,
即(n-1)•2n+1>118,
∵当n∈N+时,(n-1)•2n+1单调递增,
∴正整数n的最小取值为5.
两式相减得:(an+1-an)=2(an-an-1).
∵bn=an+1-an,
∴bn=2bn-1.
又b1=a2-a1=(2a1+1)-a1=a1+1=2.
∴数列{bn}是以2为首项,以2为公比等比数列;
(2)由(1)得bn=2n,即an+1−an=2n,
∴an=a1+(a2−a1)+(a3−a2)+…+(a n−an−1)=1+2+22+…2n−1=2n−1,
∴nan=n•2n−n,
∴Sn=(1•21−1)+(2•22−1)+…+(n•2n−n)=(1•21+2•22+…+n•2n)−
n(n+1)
2,
令T=1•21+2•22+…+n•2n ①,
则2T=1•22+2•23+…+(n-1)•2n+n•2n+1 ②,
①-②得:-T=-2+2n+1-n•2n+1,
∴T=(n-1)•2n+1+2,
∴Sn=(n−1)•2n+1+2−
n(n+1)
2,
由Sn+
n(n+1)
2>120,得(n-1)•2n+1+2>120,
即(n-1)•2n+1>118,
∵当n∈N+时,(n-1)•2n+1单调递增,
∴正整数n的最小取值为5.
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