What's the sum of $$1+\frac{1}{3}.\frac{1}{2}+\frac{2}{3}\frac{5}{6}\frac{1}{2^2}+\frac{1\cdot2\cdot5\cdot8}{3\cdot6\cdot9\cdot2^3}+\cdots$$ I think it's the expansion of some expression but can't figure it out. The answer is $\approx 4^{1/3}$ it isnt duplicate as your question is product of consecutive odd terms but mine is differenet with $2$ and difference $3$
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2What have you done so far? Have you worked out an expression for the $n^{th}$ term? Please provide more details of what you have tried and where you are stuck. This gives people a better idea of your background so they can tailor an answer to your specific level of understanding. It also shows people that you aren't just looking for someone to do your homework for you. – Ian Miller Feb 18 '16 at 08:35
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2Could you maybe also post the 4-th summand so that it becomes easier to see what's going on? – j4GGy Feb 18 '16 at 08:35
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This is what is exactly given in my book i copied the exact thing i am thinking it now for more than $10$ mins – Archis Welankar Feb 18 '16 at 08:38
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1it seems like the denominator in the $n$th term is $2^n\cdot3\cdot6\cdot9\cdots 3n=6^nn!$ and the numerator is $(3\cdot1+2)(3\cdot2+2)\cdots(3\cdot n+2)$. This however does not lead us to a sum of $4^{1/3}$ – Feb 18 '16 at 08:42
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1Possible duplicate of Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $ – lab bhattacharjee Feb 18 '16 at 09:11
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HINT...consider the binomial expansion of $(1-\frac 12)^{-\frac 23}$
David Quinn
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@ lab bhattacharjee...clearly powers of 2 in the denominator, and once you allow for the factorials, you have successive powers of 3 in the denominator, with numerators going 2, 5, 8, indicating $-\frac 23,-\frac 53, -\frac 83$. therefore $-\frac 12$ is needed to ensure all terms are positive...does this help, or do I need to write out a complete solution? – David Quinn Feb 18 '16 at 10:12