Question 1087090
<pre><font size = 4><b>
S<sub>n</sub> = 3+33+333+3333+... to n terms

If we multiply through by 3

3S<sub>n</sub> = 9+99+999+9999+... to n terms

If we add 1 to each of the n terms on the right, we will 
have added n to both sides:

3S<sub>n</sub>+n = 10+100+1000+10000+... to n terms

That is a geometric series with sum

 {{{(a[1](r^n-1))/(r-1)=(10(10^n-1))/(10-1)}}}{{{""=""}}}{{{expr(10/9)(10^n-1)}}}

So we have

{{{3S[n]+n}}}{{{""=""}}}{{{expr(10/9)(10^n-1)}}}

Solve for S<sub>n</sub>

{{{3S[n]}}}{{{""=""}}}{{{expr(10/9)(10^n-1)-n}}}

{{{S[n]}}}{{{""=""}}}{{{expr(10/27)(10^n-1)-n/3}}}

Edwin</pre></font></b>