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So I am having trouble getting the sum of the series:

$1 + 2\left(\frac32\right) + 3\left(\frac{3^2}{2^2}\right) + ... + k\left(\frac{3^{k-1}}{2^{k-1}}\right)$

I cant figure out for the life of me how to figure out the sum of this.

User
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user289882
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  • See https://en.wikipedia.org/wiki/Arithmetico-geometric_sequence#Sum_to_infinite_terms OR http://www.artofproblemsolving.com/wiki/index.php/Arithmetico-geometric_series#Infinite_Sum – lab bhattacharjee Feb 17 '16 at 08:29
  • Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you write what your thoughts are on the problem and include your efforts (work in progress) in this and future posts and in what context you have encountered the problem; this will prevent people from telling you things you already know, and help them give their answers at the right level. – JKnecht Feb 17 '16 at 08:31
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    See this. – David Mitra Feb 17 '16 at 09:24

3 Answers3

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Let $$ S = 1+a+a^2+\cdots + a^k = {1-a^{k+1}\over 1-a} $$ Differentiating for $a$ we get $$ S^\prime = 1 + 2a + 3a^2 + \cdots + ka^{k-1} = {a^k(ka-k-1)+1\over (a-1)^2} $$ It is then obvious that for $a={3\over 2}$ we get the wanted sum, that is $$ 1+2\bigl({3\over 2}\bigr) + \cdots + k\bigl({3\over 2}\bigr)^{k-1}=\bigl({3\over 2}\bigr)^k(2k-4)+4 $$

nickchalkida
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$$\sum\limits_{i=1}^{k}{i \cdot \left(\frac{3}{2}\right)^{(i-1)}}=\sum\limits_{i=0}^{k-1}{\sum\limits_{j=i}^{k-1}{{\left(\frac{3}{2}\right)}^j}}$$ $$=\sum\limits_{i=0}^{k-1}{\frac{ \left(\frac{3}{2}\right)^i -\left(\frac{3}{2}\right)^k}{1-\frac{3}{2}}}$$ $$=\frac{\frac{1- \left(\frac{3}{2}\right)^k }{ 1-\frac{3}{2}}-k\left(\frac{3}{2}\right) ^k}{-\frac{1}{2}}$$ $$=(2k-4)\left(\frac{3}{2}\right)^k+4$$

MathIsNice1729
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$$ \sum_{k=1}^{n} k a^{k-1} = \frac{d}{da}\sum_{k=1}^n a^k = \dfrac{d}{da}\left(\sum_{k=0}^na^k -1\right) $$

Chinny84
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