Question

series,
can you please help me with the first one so that I can go on to the rest by myself


https://www.dropbox.com/s/7d5qtsicifns8cu/Screenshot%202016-04-03%2012.01.18.png?dl=0

Answer 1

Preferably part b)

Answer 2

as it is my first time encountering expression to the kth power . . .

Answer 3

Terms like \(a^kb^kc^k \dotsb z^{-k}\) must be written as \((\frac{abc\dots}{z})^k\).
Then use the general formula of the geometric series: \(\sum_k x^k = 1/(1-x)\) with the new \(x=\frac{abc\dots}{z}\).

First one: \((-1)^kx^k = (-x)^k\).

Answer 4

series' value: \(\frac{1}{1 - (-x)} = \frac{1}{1+x}\).
values: term must be between -1 and 1. (theory:) \(-1<-x<1 \iff -1<x<1\) (what's given in the statement) ok!

Answer 5

well b says 2 < x < 4

Answer 6

Yes, but the general term of the series in b) is \((x-3)^k\).
-> theory wants: \(-1<(x-3)<1 \iff -2<x<4\)

Answer 7

also how is this true ... https://www.dropbox.com/s/b0dqb4tv4ojvphl/Screenshot%202016-04-03%2012.10.23.png?dl=0

Answer 8

\((ab)^\alpha = a^\alpha b^\alpha\) is true for any value of \(a,b,\alpha\).

Answer 9

so we should have 2k in the exponent

Answer 10

any real value (\(\mathbb{R}\))

Answer 11

hm..

Answer 12

ok...

Answer 13

in \((-1)^kx^k = ((-1)x)^k\), so \(a=-1,b=x,\alpha=k\).

Answer 14

one last quastion ... why is it that we have 1 / 4 - x in b) ?

Answer 15

the closed form of geometricc serie is a/1 - r right ...

Answer 16

a/(1 - r) *

Answer 17

yes

Answer 18

ok so how b) equals to 1/(4 - x) . . .

Answer 19

assuming that \(-1<x-3<1\), you use the formula \(a/(1-r)\) with \(r = x-3\). (a=1)
So
\[
\frac{1}{1- (x-3)}
\]

Answer 20

Ok thx

Answer 21

things like \(a^kb^{2k}c^{5^k}\), to be written as \((\text{term})^k\), you must use the above property of powers, and also \(x^{ab} = (x^a)^b\) it is also equal to \( (x^b)^a\).

-> \(a^kb^{2k}c^{5k} = (ab^2c^5)^k\).

useful for part c).