设数列{an}对所有正整数n都满足：a1+2a2+2^2a3+…+2^(n-1)an=8-5n,求数列{an}的前n项和

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设数列{an}对所有正整数n都满足：a1+2a2+2^2a3+…+2^(n-1)an=8-5n,求数列{an}的前n项和Sn

由：a1+2a2+2^2a3+…+2^(n-1)an=8-5n--------------------------------①
知：a1+2a2+2^2a3+…+2^(n-2)a(n-1)=8-5(n-1)-----------------------②
①-②得：2^(n-1)an=-5 n≥2
∴ an=-5*(1/2)^(n-1) n≥2
a1=8-5=3
∴S1=a1=3
Sn=a1+(a2+.+an)
=3-5×[1/2+.+(1/2)^(n-1)]
=3-5×[1-(1/2)^(n-1)] n≥2

设数列{an}对所有正整数n都满足：a1+2a2+2^2a3+…+2^(n-1)an=8-5n,求数列{an}的前n项和设数列An对所有自然数n,都满足a1+2a2+2^2a3+…+2^n-1an=8-5n,求数列a n 的通项公式,求具体设数列an满足a1+3a2+3^2a3+……+3^(n-1)an=n/3,a是正整数,设bn=n/an,求数列bn的前n 设数列{an}的前n项和为Sn,并且满足2Sn=an²+n,an>0.(1)求a1,a2,a3.(2)猜想{a 设数列{an}满足a1+3 a2+3^2 a3+……+3^n-1 an=n/3,a属于N* 求数列{an}的通项设数列an满足a1+3a2+3^2a3+.+3^n-1an=n/3,n∈N*,求数列an的通项公式已知数列an满足a1+2a2+3a3+...+nan=n(n+1)*(n+2),则数列an的前n项和Sn=? 设数列an满足a1+2a2+3a3+.+nan=2^n 1求数列a的通项 2设bn=n^2an 求数列的前n项和Sn求大设数列an的前n项和为Sn,满足2Sn=an－2∧n+1 +1 ,且a1,a2+5,a3成等差设数列{an}满足a1+3*a2+3^2*a3+......+3^(n-1)*an=3/n,n属于正整数。 (1)求数列设数列{An}满足A1+3A2+3^2*A3+...+3^(n-1)*An=n/3,a属于正整数. 数列{an}对一切自然数n属于N+满足a1+2a2+22a3+...+2n-1an=9-6n,求{an}的通项公式