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pochemuha
3 years ago
12

What is the GCF of the polynomials terms? 14a^3b^4-7ab^7+21a^2b

Mathematics
1 answer:
cupoosta [38]3 years ago
5 0

Answer:

\large\boxed{7ab}

Step-by-step explanation:

14a^3b^4-7ab^7+21a^2b\\\\14a^3b^4=2\cdot\boxed7\cdot\boxed{a}\cdot\boxed{b}\cdot b\cdot b\cdot b\\\\7ab^7=\boxed7\cdot\boxed{a}\cdot\boxed{b}\cdot  b\cdot b\cdot b\cdot b\cdot b\cdot b\\\\21a^2b=3\cdot\boxed7\cdot\boxed{a}\cdot a\cdot\boxed{b}\\\\GCF(14a^3b^4,\ -7ab^7,\ 21a^2b)=\boxed7\cdot\boxed{a}\cdot\boxed{b}=7ab

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

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\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

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s2008m [1.1K]

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