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NARA [144]
3 years ago
13

The fourth term of the sequence is 216. The sixth term is 96.

Mathematics
1 answer:
victus00 [196]3 years ago
6 0

start with 396 , then subtract 60

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What are all values of x for which the series shown converges? ​
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One convergence criteria that is useful here is that, if aₙ is the n-th term of this sequence, then we must have:

Iaₙ₊₁I < IaₙI

This means that the absolute value of the terms must decrease as n increases.

Then we must have:

\frac{(x -2)^n}{n*3^n} > \frac{(x -2 )^{n+1}}{(n + 1)*3^{n+1}}

We can write this as:

\frac{(x -2)^n}{n*3^n} > \frac{(x -2 )^{n+1}}{(n + 1)*3^{n+1}} = \frac{(x -2)^n}{(n + 1)*3^n} * \frac{(x - 2)}{3}

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\frac{(x -2)^n}{n*3^n} > \frac{(x -2)^n}{(n)*3^n} * \frac{(x - 2)}{3}

Then we have the inequality

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-1 < x < 5

The first option looks like this, but it uses the symbols ≤≥, so it is not the same as this, then the correct option will be the second.

5 0
2 years ago
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