Print the sum of the series upto n numbers

Question

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1 - (2/3!) + (3/4!) - (4/5!) + .... ± (n/(n+1)!).

What I have tried:

#include<stdio.h>
#include<conio.h>
void main()
{
int i,sum=0;
for(j=0;j<=n;j++)
{
sum+=2i/i+1;
}
printf("the sum=%d",sum);
}

Posted 25-Oct-17 7:20am

Dinesh Sahoo

Updated 25-Oct-17 18:46pm

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Comments

C^hris L^osinger 25-Oct-17 13:27pm

What you've posted here won't even compile.

Richard Deeming 25-Oct-17 13:32pm

* 2i/i + 1 would mean (2i/i) + 1, not 2i/(i+1);

* Except I don't think C lets you multiply a variable by a number without specifying the multiplication operator (2 * i);

* 2 * i/(i + 1) is nowhere near the same thing as i/(i + 1)! - multiplying by 2 is not the same as taking the factorial[^] of the divisor;

* The sequence is supposed to alternate between adding and subtracting the terms.

* And to top if off, you're not even using the loop variable!

Richard Deeming 25-Oct-17 13:35pm

You should also assume that your teacher is monitoring this site, and is aware of your attempts to cheat on your homework.

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Rick York · Answer 1 · 2017-10-25T07:52:00

Your for loop needs to have i as its variable since you have declared it and not j.

You don't have a question to answer but I'll give some advice anyway. First, write a function to compute the factorial of a value. That will provide the denominator for each value in the series. The rest of this will be a simple division of values and summing the results inside a for loop.

The only semi-tricky thing here is the alternating addition and subtraction. You can do that several ways. One is to use a multiplier and alternate it from +1 to -1. Another is to have a boolean flag that alternates from true to false and tells you whether you should add or subtract the result. Yet another is to determine addition or subtraction based on whether to loop index is odd or even which you can find taking modulo 2 of the loop index.

Rajeesh_R · Answer 2 · 2017-10-25T18:46:00

I think you should do similar like this.

C++

#include<stdio.h>
#include<conio.h>
int fact(int n)
{
    if(n == 0)
    {
       return 1;
    }
    return n * fact(n-1);
}
void main()
{
    int sum = 1, delta = -1;
    for(int i=2; i<=n; ++i)
    {
       sum += ((i / fact(i + 1)) * delta);
       delta *= -1;
    }
    printf("the sum=%d",sum);
}

I didn't compile the above code. But I think it would help you.

Print the sum of the series upto n numbers

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