Answered: "Find the series"

geduiwelh

Answered question

2021-01-05

Find the series

sweererlirumeX

Skilled2021-01-06Added 91 answers

Given series is

\sum_{n = 0}^{\infty} \frac{(x - 4)^{2 n}}{36^{n}}

Now,
let

(x - 4)^{2} = y

So, the series becomes

\sum_{n = 0}^{\infty} \frac{y^{n}}{36^{n}}

So,

a_{n} = \frac{1}{36^{n}}

Then,

(a_{n})^{\frac{1}{n}} = \frac{1}{36}

and

lim_{n \to \infty} (a_{n})^{\frac{1}{n}} = \frac{1}{36}

So, radius of convergence is 36
Then, the radius of convergence of the given series is 6
Interval of convergence

= (- 6 - 4, - 6 + 4) = (- 10, - 2)

Now,

a_{0} = 1

a_{1} = \frac{y}{36}

a_{2} = \frac{y^{2}}{36^{2}}

⋮

a_{n} = \frac{y^{n}}{36^{n}}

So,

S_{n} = a_{0} + a_{1} + . . . + a_{n}

= 1 + \frac{y}{36} + \frac{y^{2}}{36^{2}} + . . . + \frac{y^{n}}{36^{n}}

= \frac{1 [1 - (\frac{y}{36})^{n}]}{1 - \frac{y}{36}}

= \frac{36 [1 - \frac{(x - 4)^{2 n}}{36^{n}}]}{36 - (x - 4)^{2}}

= \frac{36^{n} - (x - 4)^{2 n}}{36^{n - 1} [36 - (x - 4)^{2}]}

Hence, the sum of the series as a function of x is

= \frac{36^{n} - (x - 4)^{2 n}}{36^{n - 1} [36 - (x - 4)^{2}]}

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?