The sum 1 + 2 cdot 3 + 3 cdot 3^{2} + dots + | vedclass

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(B) Let $S = 1 \cdot 3^{0} + 2 \cdot 3^{1} + 3 \cdot 3^{2} + \dots + 10 \cdot 3^{9}$.Multiplying by $3$,we get $3S = 1 \cdot 3^{1} + 2 \cdot 3^{2} + \

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$\frac{19 \cdot 3^{10} + 1}{4}$

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(B) Let $S = 1 \cdot 3^{0} + 2 \cdot 3^{1} + 3 \cdot 3^{2} + \dots + 10 \cdot 3^{9}$.Multiplying by $3$,we get $3S = 1 \cdot 3^{1} + 2 \cdot 3^{2} + \dots + 9 \cdot 3^{9} + 10 \cdot 3^{10}$.Subtracting the two equations: $S - 3S = 1 \cdot 3^{0} + (2-1) \cdot 3^{1} + (3-2) \cdot 3^{2} + \dots + (10-9) \cdot 3^{9} - 10 \cdot 3^{10}$.$-2S = (1 + 3^{1} + 3^{2} + \dots + 3^{9}) - 10 \cdot 3^{10}$.The sum in the parenthesis is a geometric progression with $a=1$,$r=3$,and $n=10$ terms.$-2S = \frac{1(3^{10} - 1)}{3 - 1} - 10 \cdot 3^{10}$.$-2S = \frac{3^{10} - 1}{2} - 10 \cdot 3^{10}$.$-2S = \frac{3^{10} - 1 - 20 \cdot 3^{10}}{2} = \frac{-19 \cdot 3^{10} - 1}{2}$.$S = \frac{19 \cdot 3^{10} + 1}{4}$.

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

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