Can anyone help me in computing this sum? $$\sum_{k=1}^{l}4^{k}3^{l+1-k}$$


closed as off-topic by José Carlos Santos, Saad, Paul Frost, Pierre-Guy Plamondon, Cesareo Jan 4 at 12:50

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  • 4
    $\begingroup$ $3^{l+1}\sum_{k=1}^l(\frac{4}{3})^k$ $\endgroup$ – W. mu Jan 4 at 8:22
  • $\begingroup$ Hi, welcome to MSE! We require questions to come with context, typically in the form of thoughts you have about the questions, attempts you've made, or even just your understanding of the concepts involved. Perhaps you could expand your question by saying why you think the terms form a geometric progression? $\endgroup$ – Theo Bendit Jan 4 at 8:22
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    $\begingroup$ Hint: Try dividing the $k+1$th term by the $k$th term. Does the result depend on $k$? (It's ok if it depends on $l$.) $\endgroup$ – Theo Bendit Jan 4 at 8:23
  • $\begingroup$ Abstract algebra is a misleading tag $\endgroup$ – roman Jan 4 at 8:24

$$ \sum_{k=1}^n 4^k3^{n+1-k} = 3^{n+1}\sum_{k=1}^n \left({4\over 3}\right)^k \\ = 3^{n+1}\left({4\over 3} + \left(4\over 3\right)^2 + \cdots + \left(4\over 3\right)^n\right) $$

Which is a regular geometric sum as you noticed: $$ \begin{align} {4\over 3} + \left(4\over 3\right)^2 + \cdots + \left(4\over 3\right)^n &= \frac{{4\over 3}\left(1-\left({4\over 3}\right)^n\right)}{1-{4\over 3}} \\ &= 4\left({4\over 3}\right)^n - 4 \end{align} $$


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