The sumn 1 + 2 * 3 + 3 * 3^2 + ... ... * 10 * 3^9n is equal to

(b) : We haven S = 1 + 2 * 3 + 3 * 3^2 + ... .. + 10 * 3^9n nn\[ n3S = 1 * 3 + 2 * 3^2 + ... ... + 9 * 3^9 + 10 * 3^10\ (i) n ]nnSubtracting (ii) from (i), we get n- 2S = ( 1 + 3 + 3^2 + ... ... + 3^9 ) - 10 * 3^10nn => - 2S = frac1( 3^10 - 1 )2 - 10 * 3^10 => 2S = frac1 - 3^102 + 10 * 3^10nn= frac1 - 3^10 + 20 * 3^102 => S = (1 / 4)lbrack 19 * 3^10 + 1 rbrack

The sum 1 + 2 \cdot 3 + 3 \cdot 3^{2} + \cdot 10 \cd... | Sequence & Series (Mathematics)

(b) : We have\n

S = 1 + 2 \cdot 3 + 3 \cdot 3^{2} + \ldots.. + 10 \cdot 3^{9}

\n\end{enumerate}\n\n\[\begin{matrix}\n3S = 1 \cdot 3 + 2 \cdot 3^{2} + \ldots\ldots + 9 \cdot 3^{9} + 10 \cdot 3^{10}\#(i) \\\n\end{matrix}\]\n\nSubtracting (ii) from (i), we get\\\n

- 2S = \left( 1 + 3 + 3^{2} + \ldots\ldots + 3^{9} \right) - 10 \cdot 3^{10}\n

\Rightarrow - 2S = \frac{1\left( 3^{10} - 1 \right)}{2} - 10 \cdot 3^{10} \Rightarrow 2S = \frac{1 - 3^{10}}{2} + 10 \cdot 3^{10}\n

= \frac{1 - 3^{10} + 20 \cdot 3^{10}}{2} \Rightarrow S = \frac{1}{4}\left\lbrack 19 \cdot 3^{10} + 1 \right\rbrack

Explanation

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