Question

Maclaurinrows - Understanding the following

Task: http://prntscr.com/3pltw4
Shown solution: http://prntscr.com/3pluf8

Translation:
c) Set up a Maclaurin row
d) Use the result in c) to decide the sum of this row

I am unable to understand the solution in d). How do you find x = -1/3? Please show me using math :)

Answer 1

@Loser66

Answer 2

first apply the Mac series representation of ln(x+1) ... and subtract x from it. this gives the numerator of your problem

then simply distribute the x^-2 to account for the denominator, the results are then worked with as needed.

Answer 3

the math is already shown for you in the solution posted .... what about it are you not grasping?

Answer 4

d)

Answer 5

\[f(x)=\sum_0\frac{(-1)^n}{n+2}x^n\]
let x = -1/3
\[f(-1/3)=\sum_0\frac{(-1)^n(\frac{-1}{3})^n}{n+2}\]
multiply thru by 3^n/3^n
\[f(-1/3)=\sum_0\frac{(-1)^n(-1)^n}{(n+2)3^n}\]
\[f(-1/3)=\sum_0\frac{(-1)^{2n}}{(n+2)3^n}\]
any even power is a postivie result soo
\[f(-1/3)=\sum_0\frac{1}{(n+2)3^n}\]
so your question is, how does this relate back to the final solution

Answer 6

"let x = -1/3"
Are you just guessing an X? How do you conclude x = -1/3, that is what I am trying to understand. :)

Answer 7

can you give a more english translation of ... bruk resultatet i c) ...

Answer 8

Translation provided in my first post

Answer 9

How do you decide to use x = -1/3?

Answer 10

The mysterious
x = -1/3... !

Answer 11

use the results from C to determine the sum of the following series ...

Answer 12

the results from c are:\[\frac{x-ln(x+1)}{x^2}=\sum_{n=0}\frac{(-1)^n}{n+2}x^n\]
D asks how we can use this knowledge to find the value of:\[\sum_{n=0}\frac{1}{(n+2)3^n}\]
in other words, solve for x

Answer 13

\[\sum_{n=0}\frac{(-1)^n}{n+2}x^n=\sum_{n=0}\frac{1}{(n+2)3^n}\]
\[\sum_{n=0}\frac{(-1)^n}{n+2}x^n-\sum_{n=0}\frac{1}{(n+2)3^n}=0\]
\[\sum_{n=0}\frac{(-1)^n}{n+2}x^n-\frac{1}{(n+2)3^n}=0\]
\[\sum_{n=0}\frac{(-1)^n3^nx^n-1}{(n+2)3^n}=0\]
if all the terms are 0, then it has to sum to zero by default
\[\sum_{n=0}\frac{(-1)^n3^nx^n-1}{(n+2)3^n}=0\]
\[\frac{(-1)^n3^nx^n-1}{(n+2)3^n}=0\]
\[(-1)^n3^nx^n-1=0\]
\[(-1)^n3^nx^n=1\]
\[x^n=\frac{1}{(-3)^n}\]
\[(x^n)^{1/n}=(\frac{1}{(-3)^n})^{1/n}\]
\[x=\frac{1}{-3}\]

Answer 14

therefore the sum is:\[\frac{(-1/3)-ln(-1/3+1)}{1/9}\]

Answer 15

I love you. Thank you!

Answer 16

lol .... i used google translate to determine what the instructions were ... after that it was pretty straight forward.