Question

已知等差数列{an}的前3项和为6，前8项和为-4

设bn=(4-an)q^(n-1) (q不等于0，n属于正整数)求数列{bn}的前n项和Sn

Answer 1

an= a1+(n-1)d
前3项和= (a1+d)3 = 6
a1+d = 2 (1)
前8项和=(2a1+7d)4 = -4
2a1+7d= -1 (2)
2(1)-(2)
-5d=5
d=-1
a1=3
an = 3-(n-1) = 4-n
bn = (4-an)q^(n-1)
= n.q^(n-1)
consider
1+x+x^2+..+x^n= [x^(n+1) -1]/(x-1)
1+2x+3x^2+..+nx^(n-1) = {[x^(n+1) -1]/(x-1)}'
= [ nx^(n+1) -(n+1)x^n + 1]/(x-1)^2
Sn = [ nq^(n+1) -(n+1)q^n + 1]/(q-1)^2

Answer 2

求啥