Question

1. calculate S = (1/1^2 + 1/2^2 + 1/3^2 + ... and so on), with 1% accuracy. Make the accuracy to be a parameter that can be assigned. For example, acc=0.01. Use "sum1" code as a template. You will need an idea for the "accuracy part" since N is not given. My suggestion: when you add a next term to S (let's say 1/n^2), check if dS/S >= acc. If 'yes' then you continue, if 'not' then you are done and can quit the loop.

program should be in c++ coding.

Answer 1

#source code:

#include<iostream>

#include<cmath>

using namespace std;

int main(){

float total=0,i=0,check;

while(1){

check=1/(pow(++i,2));

if(check>=0.01){

total+=check;

}

else{

break;

}

}

cout<<"The total value :"<<total<<"\n";

return 0;

}

#if you  have any doubt comment below..