please help to print series as well sum of series like 1*3-3*5+5*7 up to n terms i have used code like this in php
class series {
function ser(){
$i = 0;
$k = 3;
$m = 0;
for($j = 1; $j < 3; $j++) {
if($j % 2 == 0 ) {
$m = $i + ($i * $k);
} else {
$m=$m-($i*$k);
}
}
//$m = $m + $i;
echo "$m";
}
}
$se = new series();
$se->ser();
Just i have tested for 2 times
With a few simple operations one can find a formula for the sum S. If n is even (sum Se) adding pairs of terms yields
Se = (1*3 - 3*5) + (5*7 - 7*9) + (9*11 - 11*13) ...
Se = -4*( 3 + 7 + 11 + ... )
The terms in the parenthesis can be splitted and summed up:
Se = -4*( 1+2 + 3+4 + 5+6 + ... )
Se = -4*( n*(n+1)/2 )
Se = -2*n*(n+1)
If n is odd (sum So) the last term must be added to Se:
So = Se + 4*n*n-1
So = +2*n*(n+1) - 1
The implementation in C:
int series ( unsigned int n )
{
if ( n%2 == 0 )
return -2*n*(n+1);
else
return +2*n*(n+1) - 1;
}
This is probably homework, but here goes anyway. Hopefully you learn something from this.
The code above is horrible. Over-complicated for nothing... Here's a very simple version for you. I have no idea of what language this is in, but I'll do something similar for you... Go get a book on programming, that will be a wise investment of your time.
function my_sum(int $count) {
$result = 0;
$sign = 1;
for ($i=1; $i<=$count; $i++) {
$result = $result + $sign * (2*$i-1) * (2*$i+1);
$sign = - $sign;
}
return $result;
}
Hope this helps... You probably get the idea with this.
I prefer the recursive function and by this way you can stackoverflow (woot!) :) :
public static int serie(int n){
if(n<1){
return 0;
}else{
return (n%2==0?-1:1)*(4*n*n-1)+serie(n-1);
}
}
Or, use the following to compute the first n terms of your series. Sorry haven't figured out how to make SO display LaTeX properly, perhaps someone can edit it for me, but if you do please leave a comment with instructions please !
\frac{1}{2} \left(-4 (-1)^n n^2-4 (-1)^n n+(-1)^n-1\right)
Or, as generated by the wonderful EquationSheet.com:
(source: equationsheet.com)
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