# Hackerrank Calculate the Nth term Solution

Objective

A function that calls itself is known as a recursive function. The C programming language supports recursion. But while using recursion, one needs to be careful to define an exit condition from the function, otherwise it will go into an infinite loop.

To prevent infinite recursion,  statement (or similar approach) can be used where one branch makes the recursive call and other doesn't.void recurse() {    .....    recurse()  //recursive call    .....}int main() {    .....    recurse(); //function call    .....}

There is a series, , where the next term is the sum of pervious three terms. Given the first three terms of the series, , , and  respectively, you have to output the nth term of the series using recursion.

Recursive method for calculating nth term is given below.

Input Format.MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}

• The first line contains a single integer, .
• The next line contains 3 space-separated integers, , , and .

Constraints.MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}

Output Format.MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}

Print the nth term of the series, .

Sample Input 0.MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0} 51 2 3

Sample Output 011

Explanation 0.MathJax_SVG_LineBox {display: table!important} .MathJax_SVG_LineBox span {display: table-cell!important; width: 10000em!important; min-width: 0; max-width: none; padding: 0; border: 0; margin: 0}

Consider the following steps:

From steps , , , and , we can say ; then using the values from step , , , and , we get . Thus, we print  as our answer.

### Solution in c

Approach 1.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
//Complete the following function.

int find_nth_term(int n, int a, int b, int c)
{
if(n==1)
return a;
else if(n==2)
return b;
else if(n==3)
return c;
else
return (n-1)+(n-2)+(n-3);
return 0;
}

int main() {
int n, a, b, c;

scanf("%d %d %d %d", &n, &a, &b, &c);
int ans = find_nth_term(n, a, b, c);

printf("%d", ans);
return 0;
}

Approach 2.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
//Complete the following function.

int find_nth_term(int n, int a, int b, int c) {
if(n == 1)
return a;
else if (n == 2)
return b;
else if (n == 3)
return c;

return find_nth_term(n-1,a,b,c)+find_nth_term(n-2,a,b,c)+find_nth_term(n-3,a,b,c);

}

int main() {
int n, a, b, c;

scanf("%d %d %d %d", &n, &a, &b, &c);
int ans = find_nth_term(n, a, b, c);

printf("%d", ans);
return 0;
}

Approach 3.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
//Complete the following function.

int find_nth_term(int n, int a, int b, int c) {
if(n==1)
return a;
if(n==2)
return b;
if(n==3)
return c;
return find_nth_term(n-1,a,b,c)+find_nth_term(n-2,a,b,c)+find_nth_term(n-3,a,b,c);

}

int main() {
int n, a, b, c;

scanf("%d %d %d %d", &n, &a, &b, &c);
int ans = find_nth_term(n, a, b, c);

printf("%d", ans);
return 0;
}