Find the mistake. How would you solve this problem correctly? plz answer btw those a minus this is due at 11: 59

Mathematics

1 answer:

garik1379 [7]3 years ago

4 0

Answer:

The mistake is in the second "problem" or the second part

Step-by-step explanation:

solve correctly by using pemdas when there are parenthesis but only "one" number (4) and a variable (x) you multiple (bring over the 6 and times both) leaving you with 6x + 24

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Identify whether the series sigma notation infinity i=1 15(4)^i-1 is a convergent or divergent geometric series and find the sum

const2013 [10]

Answer: The correct option is

(d) This is a divergent geometric series. The sum cannot be found.

Step-by-step explanation: The given infinite geometric series is

$S=\sum_{i=1}^{\infty}15(4)^{i-1}.$

We are to identify whether the given geometric series is convergent or divergent. If convergent, we are to find the sum of the series.

We have the D' Alembert's ratio test, states as follows:

Let, $\sum_{i=1}^{\infty}a_i$ is an infinite series, with complex coefficients $a_i$ and we consider the following limit:

$L=\lim_{i\rightarrow \infty}\dfrac{a_{i+1}}{a_i}.$

Then, the series will be convergent if L < 1 and divergent if L > 1.

For the given series, we have

$a_i=15(4)^{i-1},\\\\a_{i+1}=15(4)^i.$

So, the limit is given by

$L\\\\\\=\lim_{i\rightarrow \infty}\dfrac{a_{i+1}}{a_i}\\\\\\=\lim_{i\rightarrow \infty}\dfrac{15(4)^i}{15(4)^{i-1}}\\\\\\=\lim_{i\rightarrow \infty}\dfrac{15(4)^i}{15(4)^{i}4^{-1}}\\\\\\=\dfrac{1}{4^{-1}}\\\\=4>1.$