I originally thought this Infinite Series would be fairly easy, however I can't figure out how to evaluate it. infinity ∑ (2^(j+4)) / (7^(j-3)) j=6 is there a certain formula i need to follow? Thanks.
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Originally Posted by stevo25 I originally thought this Infinite Series would be fairly easy, however I can't figure out how to evaluate it. infinity ∑ (2^(j+4)) / (7^(j-3)) j=6 is there a certain formula i need to follow? Thanks. $\displaystyle \sum_{j=6}^{\infty}p^{j+4}{q^{j-3}}=p^4 q^3\sum_{j=6}^{\infty}\left(\frac{p}{q}\right)^j$. Look familiar?
Originally Posted by stevo25 I originally thought this Infinite Series would be fairly easy, however I can't figure out how to evaluate it. infinity ∑ (2^(j+4)) / (7^(j-3)) j=6 is there a certain formula i need to follow? Thanks. $\displaystyle \sum_{j=6}^\infty \frac{2^{j+4}}{7^{j-3}} = 2^4\cdot7^3\sum_{j=6}^\infty \frac{2^{j}}{7^{j}} = 2^4\cdot7^3\sum_{j=6}^\infty\left(\frac27\right)^j $ Now we're left with a geometric series. Can you take it from here? EDIT: Ah! Drexel28 beat me to the chase!
Ah, it is pretty simple! After that you just add 6 to the j to set the index to 0 and then put (2/7)^6 outside the summation. Then you can plug it into (ar^(n+1) - a) / (r-1) formula and take the limit as it approaches infinity. Thanks!
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