I know theres a summation series for basic $a^x$. Like $2^{x-1}$ is summed by $2^x-1$. Then how would you sum $5\times2^{x-1}$?
Summation series for a $\times$ b^x
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$\begingroup$
exponential-sum
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1Hint: Distributive law – 2012-10-05
1 Answers
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You would distribute out the $5$ and get $\sum_{i=1}^x 5\cdot 2^{i-1}=5\sum_{1=1}^x 2^{i-1}=5( 2^x-1)$
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0@GerryMyerson: I thought of that after I went to bed. My only excuse is that I followed OP's notation. Fixed and thanks. – 2012-10-05