## Flowchart to find the series of number:

## Algorithm to find the series of number:

Declare i, sum, n and t;
Read n;
for i = 1 to i <= 2 * n - 1 incrementing i by 2 {
sum = sum + i * t;
If(i * t < 0)
print i * t;
else
print + i * t;
t = (-1) * t;
}
print sum;
Here in this algorithm we declare four variables as integer, n for storing the number, i for running the for loop, sum for storing the result and t for deciding if its positive(+1) or negative(-1).

Then we read the variable n and run the for loop from i=1 to n*2-1 representing the odd number like if we choose 5 then at the 5th number 9 should be there.
And incrementing i by 2 as we need odd numbers we will start form 1 and add 2 to get odd number. Then using the if statement
we will check if the number n is less than 0 or not. If yes then print as it is and if no print +i*t .We are doing this only to print the series as positive numbers do not come with + sign. For getting the sum we add sum of previous loop and odd number of this loop.
For Eg. Let's take N =3 then loop will run from 1 to 5

- In the first iteration i=1 sum =1 and we print +1 and t becomes -1
- In the second iteration i=3 sum=1-3 as t is negative from the previous loop print -3 and t becomes 1
- In the third iteration i=5 sum =1-3+5 print 5 and t now becomes -1

Now we exit the loop as we have reached i = 5 and print the sum i.e. 3

## Implementation of the Program in C:

#include
int main() {
int i, s = 0, n, t = 1;
printf("N:");
scanf("%d", & n);
for (i = 1; i <= 2 * n - 1; i += 2) {
s = s + i * t;
i * t < 0 ? printf(" %d ", i * t) : printf(" + %d ", i * t);
t = (-1) * t;
}
printf(" \nSum is %d ", s);
}
### Output of the program