Question: code class = asciimath > 1 * 3 + 2 * 3 ^ ( 2 ) + 3 * 3 ^ ( 3 ) +

code class="asciimath">1*3+2*3^(2)+3*3^(3)+cdots+n 3^(n)=(3[(2n-1)3^(n)+1])/(4)
code class = "asciimath" > 1 * 3 + 2 * 3 ^ ( 2 )

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