Question

The sum $1+2 \cdot 3+3 \cdot 3^{2}+\ldots .+10 \cdot 3^{9}$ 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)

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