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

(1) If

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

then by this rule, we have

$a_1r = (-1)^1 \implies a_1 = \boxed{-\dfrac1r}$

$a_1r+a_2r^2 = (-1)^2 \implies -1 + a_2r^2 = 1 \implies a_2 = \boxed{\dfrac2{r^2}}$

$a_1r+a_2r^2+a_3r^3 = (-1)^3 \implies -1 + 2 + a_3r^3 = -1 \implies a_3 = \boxed{-\dfrac2{r^3}}$

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

$\displaystyle \sum_{n=1}^4a_nr^n = 1 \implies -1 + 2 - 2 + a_4r^4 = 1 \implies a_4 = \dfrac2{r^4}$

$\displaystyle \sum_{n=1}^5a_nr^n = -1 \implies -1 + 2 - 2 + 2 + a_5r^5 = -1 \implies a_5 = -\dfrac2{r^5}$

$\displaystyle \sum_{n=1}^6a_nr^n = 1 \implies -1 + 2 - 2 + 2 - 2 + a_6r^6 = 1 \implies a_6 = \dfrac2{r^6}$

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

$a_n = \boxed{\dfrac{2(-1)^n}{r^n}}$

(3) We have

$a_1+a_2+\cdots+a_n +\cdots = \displaystyle \lim_{n\to\infty}\sum_{k=1}^n a_k \\\\ = \lim_{n\to\infty}\left(-\dfrac1r + 2\sum_{k=2}^n\frac{(-1)^k}{r^k}\right) \\\\ = -\dfrac1r + 2\sum_{k=2}^\infty \left(-\frac1r\right)^k$

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

$a_1+a_2+\cdots+a_n = -\dfrac1r + \dfrac2{r(r+1)} = \boxed{\dfrac{1-r}{r(1+r)}}$

Answer 2

Answer:

$good \: luck \:$