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AURORKA [14]
3 years ago
7

Given the geometric sequence where a1=-3 and the common ratio is 9 what is the domain for n

Mathematics
1 answer:
Gala2k [10]3 years ago
5 0
\bf n^{th}\textit{ term of a geometric sequence}\\\\
a_n=a_1\cdot r^{n-1}\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
r=\textit{common ratio}\\
----------\\
a_1=-3\\
r=9
\end{cases}\implies a_n=-3(9)^{n-1}

for a geometric sequence, the values "n" can take on for it to work, is usually all whole numbers, or positive integers, including 0, or you can say { x | x ∈ ℤ; x ⩾ 0 }
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Determine the horizontal, vertical, and slant asymptotes: y=x2+2x-3/x-7
densk [106]

Answer:

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Step-by-step explanation:

The vertical asymptotes are found where a denominator factor is zero (and there is no corresponding numerator factor to cancel it). Here, that is at x = 7.

There is no horizontal asymptote because the numerator is of higher degree than the denominator.

When you divide the numerator by the denominator, you get ...

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3 0
3 years ago
Find a compact form for generating functions of the sequence 1, 8,27,... , k^3
pantera1 [17]

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(if we include k=0 for a moment)

Recall that for |x|, we have

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Take the derivative to get

\displaystyle\frac1{(1-x)^2}=\sum_{k\ge0}kx^{k-1}=\frac1x\sum_{k\ge0}kx^k

\implies\dfrac x{(1-x)^2}=\displaystyle\sum_{k\ge0}kx^k

Take the derivative again:

\displaystyle\frac{(1-x)^2+2x(1-x)}{(1-x)^4}=\sum_{k\ge0}k^2x^{k-1}=\frac1x\sum_{k\ge0}k^2x^k

\implies\displaystyle\frac{x+x^2}{(1-x)^3}=\sum_{k\ge0}k^2x^k

Take the derivative one more time:

\displaystyle\frac{(1+2x)(1-x)^3+3(x+x^2)(1-x)^2}{(1-x)^6}=\sum_{k\ge0}k^3x^{k-1}=\frac1x\sum_{k\ge0}k^3x^k

\implies\displaystyle\frac{x+4x^3+x^3}{(1-x)^4}=\sum_{k\ge0}k^3x^k

so we have

\boxed{F(x)=\dfrac{x+4x^3+x^3}{(1-x)^4}}

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kumpel [21]

Answer:

x=1 or -1/2

Step-by-step explanation:

The value of x is 1 or -1/2

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