Write the term to term rule, position-to-term (nth term) and 20thterm of the sequence 3, -1 ,-5, -9, -13 ,....

Mathematics

1 answer:

timofeeve [1]2 years ago

5 0

It's a arithmetic sequence

First term=a=3
common difference=d=-1-3=-4

Nth term rule:-

$\\ \rm\Rrightarrow a_n=a+(n-1)d$

$\\ \rm\Rrightarrow a_n=3+(n-1)-4$

$\\ \rm\Rrightarrow a_n=3-4n+4$

$\\ \rm\Rrightarrow a_n=7-4n$

20th term

$\\ \rm\Rrightarrow a_{20}=7-80=-73$

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Infinite over E n=1 4(0.5)^n-1

yan [13]

Answer:

S = 8

Step-by-step explanation:

An infinite geometric series is defined as limit of partial sum of geometric sequences. Therefore, to find the infinite sum, we have to find the partial sum first then input limit approaches infinity.

However, fortunately, the infinite geometric series has already set up for you. It’s got the formula for itself which is:

$\displaystyle \large{\lim_{n\to \infty} S_n = \dfrac{a_1}{1-r}}$

We can also write in summation notation rather S-term as:

$\displaystyle \large{\sum_{n=1}^\infty a_1r^{n-1} = \dfrac{a_1}{1-r}}$

Keep in mind that these only work for when |r| < 1 or else it will diverge.

Also, how fortunately, the given summation fits the formula pattern so we do not have to do anything but simply apply the formula in.

$\displaystyle \large{\sum_{n=1}^\infty 4(0.5)^{n-1} = \dfrac{4}{1-0.5}}\\\\\displaystyle \large{\sum_{n=1}^\infty 4(0.5)^{n-1} = \dfrac{4}{0.5}}\\\\\displaystyle \large{\sum_{n=1}^\infty 4(0.5)^{n-1} = 8}$