a(n)=( 1+ 1/n- 1)*a(n- 1)+n/2^(n- 1)

Substituting the first formula, we get n+1(n-1) a (n-1)+(n+1) [1/2 (n-1)+.

And so on: Yes.

a(n+ 1)=[(n+ 1)/ 1]a 1+(n+ 1)[ 1/2+ 1/4+.......+ 1/2^n]

that is

a(n+ 1)=(n+ 1)a 1+(n+ 1){a 1[ 1-( 1/2^n)/( 1/2)]}

Simplify:

an=2n-n*[ 1/2^(n- 1)]

Because:

bn=an/n

Get:

bn=2-[ 1/2^(n- 1)]

Because:

an=2n-n*[ 1/2^(n- 1)]

So find sn

Is to find the sum of two formulas, namely 2n and n * [1/2 (n- 1)].

sn 1 = 2n[ 1+n]/2 = n(n+ 1)

sn2=[n(n+ 1)]*( 1/2)[ 1-( 1/2)^n]/( 1/2)

sn=sn 1+sn2=n(n+ 1)[2- 1/2^n]