已知数列{an}的前n项和Sn=n2+n，数列{bn}有b1=1，bnbn+1=2n（Ⅰ）求{an}的通项；（Ⅱ）若cn=anb2n，求数 已知数列{an}的前n项和为sn=n²+2n,数列...

\u5df2\u77e5\u6570\u5217{an}\u7684\u524dn\u9879\u548c\u4e3aSn\uff1dn2+1\uff0c\u6570\u5217{bn}\u6ee1\u8db3\uff1abn\uff1d2an+1\uff0c\u524dn\u9879\u548c\u4e3aTn\uff0c\u8bbeCn=T2n+1-Tn\uff0e \uff081\uff09\u6c42\u6570

\uff081\uff09a1=2\uff0c\u5f53n\uff1e1\u65f6\uff0can=Sn-Sn-1=2n-1\u2234bn\uff1d23\uff0c(n\uff1d1)1n\uff0c(n\uff1e1)\uff082\uff09Cn=T2n+1-Tn=bn+1+bn+2+\u2026+b2n+1\u2235Cn+1?Cn\uff1d12n+2+12n+3?1n+1\uff1d12n+3?12n+2\uff1c0\u2234\u6570\u5217{Cn}\u662f\u5355\u8c03\u9012\u51cf\u6570\u5217\uff0e\u7531\uff082\uff09\u77e5\uff1aCn\uff1cCn-1\uff1c\u2026\uff1cC3\uff1cC2\uff1cC1\u5f53n=1\u65f6\uff0cC1\uff1d12+13\uff1d56\uff1e1621\u5f53n=2\u65f6\uff0cC2\uff1d13+14+15\uff1d4760\uff1e1621\u5f53n=3\u65f6\uff0cC3\uff1d14+15+16+17\uff1d319420\uff1c320420\uff1d1621\u5f53n\u22653\u65f6\uff0cCn\u2264C3\uff1c1621\u6545\uff0ckmin=3\uff0e

bn=1/[(2n+1)²-1]
=1/(4n²+4n)
=1/4[1/n-1/(n+1)]
\u2234Tn
=(b1+b2+b3+\u2026\u2026+bn)
=1/4[(1-1/2)+(1/2-1/3)+(1/3-1/4)+\u2026\u2026+(1/n-1/n+1)]
=1/4[1-1/(n+1)]
=1/4[(n+1-1)/(n+1)
=n/4(n+1)

（1）当n=1时，a₁=S₁=2，
当n≥2时，a_n=S_n-S_n-1=（n²+n）-[（n-1）²+（n-1）]=2n，
又n=1时也符合上式．
∴a_n=2n．
（2）∵bnbn+1=2n，∴当n≥2时，bn-1bn=2n-1．
∴

bnbn+1

bn-1bn

=

bn+1

bn-1

=

2n

2n-1

=2，
又∵b₁=1，b₁b₂=2，∴b₂=2．
∴数列{b_2n} 是以2为首项，2为公比的等比数列，
∴b2n=2n．
∴c_n=a_nb_2n=n?2ⁿ⁺¹．
∴Tn=1×22+2×23+…+n?2ⁿ⁺¹     ①
2Tn=1×23+2×24+…+（n-1）?2ⁿ⁺¹+n?2ⁿ⁺²   ②
①-②得：-T_n=2²+2³+…+2ⁿ⁺¹-n?2ⁿ⁺²=

4(2n-1)

2-1

-n?2ⁿ⁺²=（1-n）?2ⁿ⁺²-4，
∴T_n=（n-1）?2ⁿ⁺²+4．

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