Question

啊~一道关于高中数列的问题~~(*^__^*) 嘻嘻

设数列 ｛an｝满足a1+3a2+3^2a3+………………+3^(n-1) an=n/3 n 属于N*
1 求数列 an通项公式
2 设bn=n/an 求数列bn的前N项和Sn

+3^(n-1) an这里是 加 3的 n-1次方 再乘上an

谢谢大家~~~~

Answer 1

a(1)+3a(2)+3^2a(3)+ ……………… +3^(n-2)a(n-1) +3^(n-1)a(n) = n/3
a(1)+3a(2)+3^2a(3)+ ……………… +3^(n-2)a(n-1) = (n-1)/3
相减让尺
3^(n-1)a(n) = 1/3
a(n) = 3^(-n)

b(n) = n / a(n) = n * 3^n
S(n) = ∑(k=1->n) k * 3^k
S(n)/3 = ∑(k=1->n)k * 3^(k-1)
令f(x) = ∑(k=1->n)k * x^(k-1)
∫f(x)dx = ∫∑(k=1->旦滑陪n)k * x^(k-1)dx
= ∑(k=1->n) ∫k * x^(k-1)dx
= ∑(k=1->n) ( x^k + C )
= x ( 1 - x^n ) /模蠢 ( 1 - x ) + Cn
f(x) = (∫f(x)dx))'
= ( x ( 1 - x^n ) / ( 1 - x ) + Cn )'
= ( n * x^(n+1) - (n+1) * x^n + 1 ) / ( x - 1 )^2
S(n) = 3 * S(n)/3 = 3 * f(3)
= 3 * ( n * 3^(n+1) - ( n + 1 ) * 3^n + 1 ) / 4
= ( 2n - 1 ) * 3^(n+1) / 4 + 3/4