How do you write a program to scan a number n and then output the sum of the powers from 1 to n using for loop statement?

Answer 1

#includ<iostream.h>

int main ()

{

int n,n-1;

for(n=0);

cout<<"the number is even:"<<endl;

if(n=n-1)

cout<<"the number is prime:"<<endl;

getch 0;

return 0;

}

Answer 2

#include

using std::cin;
using std::cout;
using std::endl;

long powerOfN(long parameter, short degree);
long sumOfN(long parameter);

int main()
{
long data = 0L;
cout << endl << "Enter N: ";
cin >> data;

cout << "Result is: " << sumOfN(data) << endl;

system("PAUSE");
return 0;

}

long powerOfN(long parameter, short degree)
{
long result = 1L;
for (int i = 1; i <= degree; i++)
{
result *= parameter;

}

return result;

}

long sumOfN(long parameter)
{
long sum = 0;
for (int i = 1; i <= parameter; i++)
{
sum += powerOfN(i, 2);

}

return sum;

}

Answer 3

This only works in Xcode Command Line Tool.

#include

int main (int argc, const char * argv[])

{

int sum=0, accum1, accum2;

int i, c, n;

printf("Enter Number: ");

scanf("%d", &n);

for(i=1;c<=n;i++){

accum1=1;

accum2=i;

for(c=1;c<=i;c++){

accum1 = accum1 * accum2;

}

sum = sum + accum1;

}

printf("Sum is %d.",sum);

}

Answer 4

It is not clear what is being asked here because the powers from one to n depend on which values you apply those powers to. Assuming you mean the sum of n^1 ... n^n, then simply sum all the integers from 1 to n then raise n to that power:

#include<stdio.h>

#include<math.h> // pow() function

unsigned sum_powers (const unsigned n) {

unsigned tmp = n, pwr = 0;

while (tmp) pwr += tmp--;

return pow (n, pwr);

}

int main (void) {

unsigned n;

printf ("Enter a number: ");

scanf ("%u", n);

printf ("Sum of n^1 to n^n: %u\n", sum_powers (n));

return 0;

}

Answer 5

You cannot sum powers. A power is an operator and, like all operators, it requires an operand. That is; 2^3 means 2 raised to the power of 3, where 2 is the operand and "raised to the power of 3" is the operator. But without the operand, "raised to power of three" becomes meaningless.

To answer the question you need two pieces of information: the range of powers (n) and the value to be raised by those powers (v).

#include<math.h> // for C standard library pow() function

unsigned sum_powers (unsigned v, unsigned n) {

unsigned sum;

sum = v; // v to the power 1 is v

while (n>1) { // sum v^n for all n>1

sum = sum + pow (v, n);

n--;

}

return sum;

}

int main (void) {

unsigned n, v;

printf ("Enter a value:\n");

scanf ("%d", &v);

printf ("Enter the maximum power:\n");

scanf ("%d", &n);

printf ("The sum of %d^1 to %d^n is: %d\n", v, v, n, sum_powers (v, n));

return 0;

}