Find the series in C programming - c

Question

Help me to find the sum of this series x^1+x^4+x^7+.. to n terms
#include<conio.h>
#include<math.h>
int main(int argc, char const *argv[]) {
int n;
float x;
int l=1;
printf(" Enter the value of x :");
scanf("%f",&x);
for (int i = 1; i <=n; i++)
{
l = pow(x,l+3);
}
printf("x ^ %d + power %d",x,l);
return 0;
}

Answer 1

I think this is what you want :) i've commented the errors
#include<conio.h>
#include<math.h>
int main(int argc, char const *argv[]) {
int n;
float x;
float l=0,i=0;
// initialize n first to determine how far the sum should go
printf(" Enter the value of n :");
scanf("%d",&n);
printf(" Enter the value of x :");
scanf("%f",&x);
for (int i = 0; i <n; i++)
{ //l+= pow(x,(i*3)+1); is equivilant to l= l+ pow(x,i+3); which i assume that's what you wanted to do
// l should be a float as well not int since x is a float
l+= pow(x,(i*3)+1);
if(i!=n-1)printf(" %f^%d +",x,i*3+1);
else printf("%f^%d",x,i*3+1);
}
printf("\nResult of the sum is %f",l);
return 0;
}

Answer 2

There's a problem in this loop:
for (int i = 1; i <=n; i++)
{
l = pow(x,l+3);
}
You set l every time and then in the end it has the value of the last assignment, but you wanted to add them all up. Also, it should be i in the pow call, not l+3, instead you should move that step of 3 to the loop increment. Try this:
int l = 0;
for (int i = 1; i <= n; i+=3)
{
l += pow(x, i);
}
Also make sure that you initialize n, you should probably read it in using scanf, just like you do with x.
Also note that since l is int, it's going to be truncated every time. If you don't want that, you should declare it as float instead, and depending on how precise the result should be, you might even want to consider double.

Answer 3

#include <stdio.h>
#include <math.h>
int main()
{
double x;
unsigned n;
fprintf(stderr, "enter x and n :");
/* enter x as a double and x as an unsigned number, both on the same line, and check 2 values are enter */
if (scanf("%lg %u", &x, &n) != 2)
puts("invalid or missing values");
else {
double r = 0; /* sum initialization */
int p = 1; /* the power to apply, 1 then 4 then 7 etc */
while (n--) { /* have to sum n times */
r += pow(x, p); /* x^1+x^4+x^7+... */
p += 3; /* next power value */
}
printf("%g\n", r);
}
return 0;
}
if I well understand n terms x^y must be added rather than the last pow is x^n, I use double for x and of course the result ( r )
Executions :
enter x and n :1.2 3
6.85678
enter x and n :2 1
2
Check with bc :
% bc
bc 1.06.95
Copyright 1991-1994, 1997, 1998, 2000, 2004, 2006 Free Software Foundation, Inc.
This is free software with ABSOLUTELY NO WARRANTY.
For details type `warranty'.
scale=6
1.2 + 1.2^4 + 1.2^7
6.856780

Answer 4

Multiple operations on floating point numbers tend to compound rounding errors very quickly, so if you can minimize the number of operations, always do.
Now if you recognize that your polynomial is a Geometric series (see wikipedia article) with a = x and r = x^3, then we have the following pseudo-code algorithm (note no loops!):
Case 1 (x == 0): Sum == 0 regardless of n.
Case 2 (x == 1): Sum == n.
Case 3 (x any other number): Sum == x*(1-(x^(3*n)))/(1-(x^3)).
Case 4 (n==infinity and -1<x<1): Sum == x/(1-(x^3)).
I trust you can code accordingly.