求数列2+a,5+a²,8+a³,.(3n-1)+a^n的前几项和
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求数列2+a,5+a²,8+a³,.(3n-1)+a^n的前几项和
求详细解答过程谢谢
求详细解答过程谢谢
let
S = 1.a^1+2.a^2+.+n.a^n (1)
aS = 1.a^2+2.a^3+.+n.a^(n+1) (2)
(2)-(1)
(a-1)S = n.a^(n+1) - ( a+a^2+...+a^n)
= n.a^(n+1) - a( a^n-1)/(a-1)
S = [1/(a-1)][n.a^(n+1) - a( a^n-1)/(a-1)]
bn= 3n-1
cn = a^n
dn=bn + cn
= (3n-1) .a^n
= 3( n.a^n) - a^n
Sn =d1+d2+...+dn
= 3S - a(a^n-1)/(a-1)
=[3/(a-1)][n.a^(n+1) - a( a^n-1)/(a-1)] - a(a^n-1)/(a-1)
数列2+a,5+a²,8+a³,.(3n-1)+a^n的前几项和
= Sn
=[3/(a-1)][n.a^(n+1) - a( a^n-1)/(a-1)] - a(a^n-1)/(a-1)
S = 1.a^1+2.a^2+.+n.a^n (1)
aS = 1.a^2+2.a^3+.+n.a^(n+1) (2)
(2)-(1)
(a-1)S = n.a^(n+1) - ( a+a^2+...+a^n)
= n.a^(n+1) - a( a^n-1)/(a-1)
S = [1/(a-1)][n.a^(n+1) - a( a^n-1)/(a-1)]
bn= 3n-1
cn = a^n
dn=bn + cn
= (3n-1) .a^n
= 3( n.a^n) - a^n
Sn =d1+d2+...+dn
= 3S - a(a^n-1)/(a-1)
=[3/(a-1)][n.a^(n+1) - a( a^n-1)/(a-1)] - a(a^n-1)/(a-1)
数列2+a,5+a²,8+a³,.(3n-1)+a^n的前几项和
= Sn
=[3/(a-1)][n.a^(n+1) - a( a^n-1)/(a-1)] - a(a^n-1)/(a-1)
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