PLZ CAN SOMEONE HELP ME!!!!!!!

Mathematics

1 answer:

Hunter-Best [27]2 years ago

5 0

The answer is 4/16. The fastest way to solve these sorts of equations is to remember that a fraction is merely a division problem. Take the top number (numerator) and divide it by the bottom number ( denominator). That will give you a decimal. 4/16 is equal to .25. 5/20 is also equal to .25. You could also go about it by reducing the fraction. 5 goes into 20 4 times 5 goes into itself 1 time. so you have 1/4. 4 goes into 16 four times and into itself once. It is also 1/4. I hope this helps. :)

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Find the radius of convergence, r, of the series. ? n2xn 7 · 14 · 21 · ? · (7n) n = 1

defon

I'm guessing the series is supposed to be

$\displaystyle\sum_{n=1}^\infty\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}$

By the ratio test, the series converges if the following limit is less than 1.

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7\cdot14\cdot21\cdot\cdots\cdot(7n)\cdot(7(n+1))}}{\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}}\right|$

The first $n$ terms in the numerator's denominator cancel with the denominator's denominator:

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7(n+1)}}{n^2x^n}\right|$

$|x^n|$ also cancels out and the remaining factor of $|x|$ can be pulled out of the limit (as it doesn't depend on $n$ ).

$\displaystyle|x|\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2}{7(n+1)}}{n^2}\right|=|x|\lim_{n\to\infty}\frac{|n+1|}{7n^2}=0$

which means the series converges everywhere (independently of $x$ ), and so the radius of convergence is infinite.

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Dima020 [189]

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