求和,1＋4／5＋7／5²＋…＋3n－2／5^（n－1）
2个回答

令Sn=1+4/5+7/5^2+…+3n-2/5^(n-1)

5Sn=5+4+7/5+10/5^2+…+3n-2/5^(n-2)

两式相减：

4Sn=8+3*(1/5+1/5^2+……+1/5^n-2)-(3n-2)/5^(n-1)

4Sn=8+3*1/5*[1-(1/5)^(n-2)]/(1-1/5)-(3n-2)/5^(n-1)

4Sn=8+3*1/5*[1-(1/5)^(n-2)]/(4/5)-(3n-2)/5^(n-1)

4Sn=8+3*[1-(1/5)^(n-2)]/4-(3n-2)/5^(n-1)

4Sn=8+3/4-3*(1/5)^(n-2)/4-(3n-2)/5^(n-1)

4Sn=8+3/4-3*(1/5)^(n-2)/4-(3n-2)*(1/5)^(n-1)

4Sn=8+3/4-3*(1/5)^(n-2)/4-(3n-2)*(1/5)^(n-2)*(1/5)

4Sn=8+3/4-[3/4+(3n-2)*(1/5)]*(1/5)^(n-2)

4Sn=8+3/4-[3/4+3n/5-2/5]*(1/5)^(n-2)

4Sn=35/4-[7/20+3n/5]*(1/5)^(n-2)

4Sn=35/4-(7+12n)*(1/5)^(n-2)/20

Sn=35/16-(12n+7)*(1/5)^(n-2)/80

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