Question

Given 'n' is a positive integer and 'r' is a real number which r > 1

Given the sequence {a_n} corresponding to:

`\displaystyle \large{a_1r + a_2 {r}^(2) + ... + a_n {r}^(n) = {( - 1)}^(n) }`
Answer the following questions:
( 1 )
Write a1,a2,a3 in term of r.
( 2 )
When n ≥ 2, write a_n in term of r.
( 3 )
Evaluate the sum of Infinite Series a1+a2+...+an+...
Thanks in advance!​

Answer 1

`good \: luck \: \\ \\ \\`

Answer 2

(1) If

`a_1r+a_2r^2+a_3r^3+\cdots+a_nr^n = (-1)^n`

then by this rule, we have

`a_1r = (-1)^1 \implies a_1 = \boxed{-\frac1r}`

`a_1r+a_2r^2 = (-1)^2 \implies -1 + a_2r^2 = 1 \implies a_2 = \boxed{\frac2{r^2}}`

`a_1r+a_2r^2+a_3r^3 = (-1)^3 \implies -1 + 2 + a_3r^3 = -1 \implies a_3 = \boxed{-\frac2{r^3}}`

(2) The next 3 terms in the sequence
`\{a_n\}` would be

`\displaystyle \sum_(n=1)^4a_nr^n = 1 \implies -1 + 2 - 2 + a_4r^4 = 1 \implies a_4 = \frac2{r^4}`

`\displaystyle \sum_(n=1)^5a_nr^n = -1 \implies -1 + 2 - 2 + 2 + a_5r^5 = -1 \implies a_5 = -\frac2{r^5}`

`\displaystyle \sum_(n=1)^6a_nr^n = 1 \implies -1 + 2 - 2 + 2 - 2 + a_6r^6 = 1 \implies a_6 = \frac2{r^6}`

The pattern should be clear; we have for n ≥ 2,

`a_n = \boxed{(2(-1)^n)/(r^n)}`

(3) We have

`a_1+a_2+\cdots+a_n +\cdots = \displaystyle \lim_(n\to\infty)\sum_(k=1)^n a_k \\\\ = \lim_(n\to\infty)\left(-\frac1r + 2\sum_(k=2)^n((-1)^k)/(r^k)\right) \\\\ = -\frac1r + 2\sum_(k=2)^\infty \left(-\frac1r\right)^k`

As long as |-1/r | < 1, or |r | > 1, the remaining geometric series converges to

`\displaystyle \sum_(k=2)^\infty \left(-\frac1r\right)^k = \sum_(k=0)^\infty\left(-\frac1r\right)^k - 1 - \left(-\frac1r\right) = \frac1{1-\left(-\frac1r\right)}-1+\frac1r = \frac1{r(r+1)}`

Then

`a_1+a_2+\cdots+a_n = -\frac1r + \frac2{r(r+1)} = \boxed{(1-r)/(r(1+r))}`