Let's assume that you want the sum of the general harmonic series:
sum(n=0,inf): 1/(an+b)
Since we know that the harmonic series will converge to infinity, we'll also assume that you want the sum from 0 to n.
double genHarmonic(const double n, const double a, const double b) {
double sum = 0.0;
// perform calculations
int k;
for(k = 0; k <= n; ++k) {
sum += 1.0 / (a * k + b);
}
return sum;
}
Write a program to find the number and sum of all integers from 100 to 300 that are divisible by 11
sum = 0; for (int i = 12; i
As you can see, this series is the sum of the sequence: 1, 2, 3, 4, 5, 6. That is to say: 1 + 2 = 3 3 + 3 = 4 6 + 4 = 10 10 + 5 = 15 15 + 6 = 21 To program this in C, you would want a simple incremental counter which would be added to the series sum on each iteration of the loop. { int c = 2; int sum = 1; // loop until you find all the numbers you want while(1) { // increase the running sum to find the next next number in the series sum += c; // increment our counter ++c; } }
The name of the program. For example: program sum ! This is a comment. Your program's code goes here... end program sum
int sum (int min, int max) {return (max-min+1)*(max+min)/2;}
Writing a program for a sum of sine series requires a rather long formula. That formula is: #include #include #include main() { int i,n,x; .
What is the assembly program to generate a geometric series and compute its sum The inputs are the base root and the length of the series The outputs are the series elements and their sum?
It depends on the series.
This series is known as the Harmonic Series and it diverges but very, very slowly. For example, the first 100 terms sum to 5.187...., the first 1000 terms to 7.486...., and the first 1000000 terms to 14.392.... There are many proofs of the divergence of this series and an internet search of Harmonic Series will no doubt find many of them.
write an assembly language program to find sum of N numbers
int sum = 0; int n = 0; while( sum <= 999 ) { sum += (++n); }
Write a program to find the number and sum of all integers from 100 to 300 that are divisible by 11
sum = 0; for (int i = 12; i
The Nth partial sum is the sum of the first n terms in an infinite series.
matrix
sample program in sum of the series using the formula for s=n/2[2a+{n-1}d] in 8085
Did you know that memory allocation is not needed to display the matrix? However, the C program is to find the sum of all the elements.