Fuzzy Operators

$S - 1=2-3\times 4\div 5\ + 6-7\times 8\div 9+ ...$

$S - 1 = n - \frac{(n+1)\times(n+2)}{n+3} + ... = n - (n + \frac{2}{n+3})+... = - \frac{2}{n+3} - ... - \frac{2}{N+4}$

$n= N+1$

$n =\left\{ 2, 6, 10, ...,N+1 \right\}$

$S = 1 - 2\times (\frac{1}{n+3} + ... +\frac{1}{N+4}$ ) = $1 - 2 \times \sum_{n=1}^\frac{N+1}{4} \frac{1}{4n+1}$

MATLAB scripts

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N = 31417;
sum = 1;

for n= 2:4:N+1
    sum = sum - 2/(n+3);
end
sum

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N = 31417;
sum = 1;

for n = 1:(N+1)/4
    sum = sum - 2/(4.*n +1);
end
sum

With Excel of cells:

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1, 1, =B1
=4*(ROW()-2)+2, =A2-(A2+1)*(A2+2)/(A2+3), =C1+B2
...

for

1 1 1   1
2 2 -0.4    0.6
3 6 -0.222222222    0.377777778
4 10    -0.153846154    0.223931624
...
7853 31406  -6.3676E-05     -3.598036301
7854 31410  -6.36679E-05    -3.598099969
7855 31414  -6.36598E-05    -3.598163629

We can check sum of terms with step accumulations to agree each other.

-3.59816362860588 is the numerical value obtained.

Answer: $\boxed{-3.598}$

There are thousands of Codes to solve this.

You can identify the Reappeared Sequence of the operators. $s=1+(2-3*4/5)+(6-7*8/9)+(10-11*12/13)$

Now, get ready for the Loop. As you see, that "1" is kept out of the braces, you have to set the initial value of Sum, s=1; .

Then you take the 4 variables - a,b,c,d. You see, always, $b=a+1$ $c=b+1$ $d=c+1$

And it occurs again and again. In the 1st loop, a=2 . You can declare it inside the for loop. To get the numbers of the 2nd loop, a+=4 . Because then you get $a=6$ in the 2nd, and $a=10$ in the 3rd loop and blah blah blah!

So, the for loop is for(a=2; ; a+=4) . Look carefully, we didn't specified the condition for the loop, because using the break; statement, we can do that job.

Then, inside the loop, first produce the other numbers, $b,c,d$ .

Now comes the most important. If $a=n$ , $s+=a;$ then Break the loop.

But if $b=n$ , then $s+=a-b;$ , and Break ! Similar for $c=n$ and $d=n$ . Notice their statements in the program below.

If nothing of them isn't equals to $n$ , that means that they haven't reached the value of $n$ yet, which will happen from the beginning. For that, the statement is, s+=a-((b*c)/d); . Remember no Break this time, as it needs to continue until any of the variables reach the $n$ .

So, that's it. Here's the code after all this detailed explanation...

Compile, run it, Enter n=31417 and get the answer!

Any queries, comment box is open for you of course.

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#include<conio.h>
#include<math.h>
int main()
{
    double a, b, c, d, s, n;
    printf("What's the n: ");
    scanf("%lf", &n);

    s = 1;
    for (a = 2;; a += 4)
    {
        b = a + 1;
        c = b + 1;
        d = c + 1;
        if (a == n)
        {
            s += a;
            break;
        }
        else if (b == n)
        {
            s += (a - b);
            break;
        }
        else if (c == n)
        {
            s += a - (b * c);
            break;
        }
        else if (d == n)
        {
            s += a - ((b * c) / d);
            break;
        }
        else

        {
            s += a - (b * c / d);
        }

    }
    printf("The Summation,s= %lf", s);
    printf("\n");


    return 0;
}

p=31417

x=2

y=3

s1=0.00

s2=0.00

while x<=p:

s1=s1+x

x=x+4.00

while y<=p:

if y+2<=p:

    s2=0.00+s2+y*(y+1)/(y+2.00)


elif p-y==1:

    s2=0.00+s2+y*(y+1)

elif p-y==0:

     s2=0.00+s2+y


y=y+4.00

print s1+1-s2