已知公差不为零的等差数列{an}的前4项和为10，且a2，a3，a7成等比数列．（Ⅰ）求通项公式an（Ⅱ）设bn=2an，求数列{bn}的前n项和Sn．

Question

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n2an，求数列{b_n}的前n项和S_n．2an，求数列{b_n}的前n项和S_n．2an，求数列{b_n}的前n项和S_n．an，求数列{b_n}的前n项和S_n．an，求数列{b_n}的前n项和S_n．an，求数列{b_n}的前n项和S_n．n，求数列{b_n}的前n项和S_n．_nn

Answer 1

（I）由题意可得，

4a1+6d＝10

(a1+2d)2＝(a1+d)(a1+6d)

∵d≠0
∴

a1＝−2

d＝3

∴a_n=3n-5
（II）∵b_n=2an=2^3n-5=(1)/(4)•8n−1
∴数列{a_n}是以(1)/(4)为首项，以8为公比的等比数列
∴Sn＝((1)/(4)(1−8n))/(1−8)=(8n−1)/(28)

4a1+6d＝10

(a1+2d)2＝(a1+d)(a1+6d)

4a1+6d＝10

(a1+2d)2＝(a1+d)(a1+6d)

4a1+6d＝10

(a1+2d)2＝(a1+d)(a1+6d)

4a1+6d＝10

(a1+2d)2＝(a1+d)(a1+6d)

4a1+6d＝104a1+6d＝104a1+6d＝101+6d＝10(a1+2d)2＝(a1+d)(a1+6d)(a1+2d)2＝(a1+d)(a1+6d)(a1+2d)2＝(a1+d)(a1+6d)1+2d)2＝(a1+d)(a1+6d)2＝(a1+d)(a1+6d)1+d)(a1+6d)1+6d)
∵d≠0
∴

a1＝−2

d＝3

∴a_n=3n-5
（II）∵b_n=2an=2^3n-5=(1)/(4)•8n−1
∴数列{a_n}是以(1)/(4)为首项，以8为公比的等比数列
∴Sn＝((1)/(4)(1−8n))/(1−8)=(8n−1)/(28)

a1＝−2

d＝3

a1＝−2

d＝3

a1＝−2

d＝3

a1＝−2

d＝3

a1＝−2a1＝−2a1＝−21＝−2d＝3d＝3d＝3
∴a_nn=3n-5
（II）∵b_nn=2an=2^3n-5=(1)/(4)•8n−1
∴数列{a_n}是以(1)/(4)为首项，以8为公比的等比数列
∴Sn＝((1)/(4)(1−8n))/(1−8)=(8n−1)/(28) 2an=2^3n-5=(1)/(4)•8n−1
∴数列{a_n}是以(1)/(4)为首项，以8为公比的等比数列
∴Sn＝((1)/(4)(1−8n))/(1−8)=(8n−1)/(28) an=2^3n-5=(1)/(4)•8n−1
∴数列{a_n}是以(1)/(4)为首项，以8为公比的等比数列
∴Sn＝((1)/(4)(1−8n))/(1−8)=(8n−1)/(28) n=2^3n-53n-5=(1)/(4)•8n−1
∴数列{a_n}是以(1)/(4)为首项，以8为公比的等比数列
∴Sn＝((1)/(4)(1−8n))/(1−8)=(8n−1)/(28) (1)/(4)111444•8n−1
∴数列{a_n}是以(1)/(4)为首项，以8为公比的等比数列
∴Sn＝((1)/(4)(1−8n))/(1−8)=(8n−1)/(28) n−1
∴数列{a_nn}是以(1)/(4)为首项，以8为公比的等比数列
∴Sn＝((1)/(4)(1−8n))/(1−8)=(8n−1)/(28) (1)/(4)111444为首项，以8为公比的等比数列
∴Sn＝((1)/(4)(1−8n))/(1−8)=(8n−1)/(28) Sn＝((1)/(4)(1−8n))/(1−8)=(8n−1)/(28) n＝((1)/(4)(1−8n))/(1−8)(1)/(4)(1−8n)(1)/(4)(1−8n)(1)/(4)111444(1−8n)n)1−81−81−8=(8n−1)/(28) (8n−1)/(28)8n−18n−18n−1n−1282828