Question

How many terms of the series $1+3+3^2+3^3+...+3^{n-1}$ must be taken to make $3280$.

Answer 1

$1+3+3^2+.....{.3}^{n-1}=3280$

GP with $a=1$ and $r=3$

Sum of n terms in GP$=\frac{a(r^n-1)}{(r-1)}$

$\therefore$ Sum of $(n-1)$ terms $\Rightarrow \frac{1(3^n-1)}{(3-1)}=3280$

$\Rightarrow 3^{\left(n\right)}-1=3280\times 2=6560$

$\Rightarrow 3^n=6561$

$\Rightarrow n=8$

$\therefore$ We need $8$ terms of the series $1+3+3^2+....$ to make $3280$.