In $C$ programming$($as a beginner$),$ write an algorithm in the main$()$ that generates $N$ first

elements of a series $s_{n_{(x)}}=-\frac{(x{)}^n}{n}.$

This series represents the Taylor series : $ln(1-x)$~~$-{\displaystyle \underset{n=1}{\overset{N=6}{}}}\frac{(x{)}^n}{n}.$

Evaluate this series in for$()$ loop for $x=0\mathrm{.}5,$ and $n=1,2,$dots,$6.$ Calculate and print the subsequent

element values in a terminal. Hint: the first three elements are $-x,\frac{-x^2}{n},\frac{-x^3}{n}.$

Modify the main $()$ such that the user can insert through a keyboard value of $m.$ Allocate

dynamically two $1-$dimensional arrays $($VecA and VecB$)$ that contain $m=10$ elements of

double type.

Write a separate function funFill$()$ that will fill a dynamically allocated vector VecA from

point $2$ with the first $10$ elements of a series defined in point $1(n=1,2,3,$dots,$10).$ Print in the

terminal values of each element in VecA

Write a separate function funLn$()$ that will fill a dynamically allocated vector VecB from point

$2$ with the sum of the first $m$ elements of a series defined in point $1($i$.$e$.$ for $n=1$ sum of the

first element, for $n=2$ sum of first two elements, for $n=3$ the sum of first three elements,etc.$)$

Print in the terminal values of each element in VecB; $n$ must be implemented as an input

argument.

Modify the funLn$()$ such that the last element of VecB is also returned into the main $()$ using

a pointer.

Plot the value of the last element of VecB for $x=0\mathrm{.}5$ in for$()$ loop for $n=1,$dots$20.$ Plot the

absolute value of the difference between the value of the last element of VecB and function

$ln(1-x).$ Save this data in a separate txt file in three columns: $n-$ first column, last element of

VecB $-$ second column, the absolute value of the difference between last element of VecB

and $ln(1-x)-$ third column.