Question

数学考试。 急

Answer 1

1+d=3/a
d=2/a
=>1+2/a=3/a
=>1=1/a
=>a=1 d=2
(1)Sn = (1+1+2(n-1))*n/2 = n^2
(2)Tn=3^0*(1)+3^1*(3) + 3^2*(5)+...+3^(n-1)*(2n-1) (1)
3Tn= (0)+3^1*(1) + 3^2*(3)+...+3^(n-1)*(2n-3)+3^n*(2n-1) (2)
(2)-(1)=>
2Tn = 3^1*2+3^2*2+...+...3^(n-1)*2+3^n*(2n-1)-1
=2(3-3^n)/(1-3)+3^n*(2n-1)-1 = 3^n-3+(2n-1)*3^n-1=2n*3^n-4
=>Tn = n*3^n-2

(3)Cn = 2n-1+1=2n
Tn = (2+2n)*n/2=n^2+n
1/Tn = 1/(n^2+n) = 1/n-1/(n+1)
1/T1+1/T2+...+1/Tn
=1/1-1/2+1/2-1/3+...+1/n-1/(n+1) = 1-1/(n+1)=n/(n+1)

Answer 2

通过方程可知道a=1,d=2
1、an=2n-1,Sn=n的平方