Question

$1+3^1+3^2+3^3..........+3^{2999}$ =

• A. $\frac{5^{2999}-1}{2}$
• B. $\frac{3^{3000}-1}{2}$
• C. $\frac{3^{2999}-1}{2}$
• D. $\frac{4^{1999}-1}{2}$

Answer 1

The correct option is B

$\frac{3^{3000}-1}{2}$

The given series is a GP with first term $a=1$ and the common ratio $r=\frac{t_2}{t_1}=\frac{3}{1}=3.$

The sum to $n$ terms of the GP is given by
$S_n=\frac{a(r^n-1)}{r-1}\text{, where}r>1.$

$\therefore 1+3^1+3^2+3^3..........+3^{2999}=\frac{1(3^{3000}-1)}{3-1}$

$=\frac{(3^{3000}-1)}{2}$