The rth term of a series is given by
$\displaystyle
u_r=(\frac{1}{3})^{3r-2}+(\frac{1}{3})^{3r-1}
$
Express $\displaystyle \sum^{n}_{r=1}u_r$ in the form
$\displaystyle A(1-\frac{B}{27^n}) $, where A and B are constants to be found ,
$\displaystyle
u_r=(\frac{1}{3})^{3r-1}\frac{3}{1}+(\frac{1}{3})^{3r-1}
$
$\displaystyle
u_r=(\frac{1}{3})^{3r-1}(3+1)
$
$\displaystyle
u_r=(\frac{1}{3})^{3r-1}(4) ~= 12\frac{1}{3^{3r}} ~= \frac{12}{27^r}
$
No need to tell you that this is a GP
$\displaystyle Sum = a\frac{(r^n-1)}{r-1}$
$\displaystyle =\frac{4}{9} \frac{(1-\frac{1}{27^n})}{(1-\frac{1}{27})}
$
$\displaystyle =\frac{4 \times 27}{9\times 26} {(1-\frac{1}{27^n})}
$
I think now you can continue to