2s (n+1) = 3a (n+1)-4 = 2 (sn+a (n+1)), which can be simplified.

An (n+1)-1= 3 (An- 1), a1-1= 2-1=1,then an-/kloc.

an- 1= 1*3^(n- 1)

2), bn = n/(3 (n- 1)), and the sum is calculated by dislocation subtraction.

Sn= 1/3+2/9+3/27…+n/3^n

Sn/3 =1/9+2/27 ...+(n-1)/3n+n/3 (n+1).

2sn/3= 1/2- 1/(2*3^n)-n/3^(n+ 1)→sn=3/4- 1/[4*3^(n- 1)]-n/2*3^n

17, 1), to find the general term: just use arithmetic progression's general formula: an=a 1+(n- 1)d to solve it.

an= 1+3(n- 1)=3n-2

2), bn =1/(3n-2) * (3n+1) can be factorized:

BN =1/3 *1/(3n-2)-1/(3n+1), so the sum is very simple.

sn = 1/3 *[ 1- 1/4+ 1/4- 1/7+.............+ 1/(3n-5)- 1/(3n-2)+ 1/(3n-2)- 1/(3n+ 1)]

= 1/3 *[ 1- 1/(3n+ 1)]

=n/(3n+ 1)

All hands, so tired!