Question

The sum $1+2\cdot 3+3\cdot 3^2+\dots .+10\cdot 3^9$ is equal to

• A. $\frac{2\cdot 3^{12}+10}{4}$
• B. $\frac{19\cdot 3^{10}+1}{4}$
• C. $5\cdot 3^{10}-2$
• D. $\frac{9\cdot 3^{10}+1}{2}$

Answer 1

Answer: (B)

$S=1\cdot 3^0+2\cdot 3^1+3\cdot 3^2+\dots \dots +10.3^9$
$3S=1\cdot 3^1+2.3^2\dots \dots \dots \dots \dots \dots \dots +9\times 3^9+10\times 3^{10}$
$-2S=\left(1\cdot 3^0+3^1+3^2\dots .3^9\right)-10.3^{10}$
$S=5\times 3^{10}-\left(\frac{3^{10}-1}{4}\right)$
$S=\frac{20.3^{10}-3^{10}+1}{4}=\frac{19.3^{10}+1}{4}$