Answer / sanath

#include <iostream.h>

void main()
{
int nTerm;
int x=1,y=3,z=5,mul1,total=0;

cout<<"Enter the NTerm:";
cin>>nTerm;
for(int i=1;i<=nTerm;i++)
{
mul1 = ((x++)*(y++)*(z++));

if(i%2)
total = total + mul1;
else
total = total - mul1;

}
cout<<"RESULT:"<<total<<endl;
}

Answer / sri k. hari

/* Author : Sri K. Hari
Description : program for series
1*3*5-2*4*6+3*5*7-4*6*8...........upto n terms
*/

#include<stdio.h>
#include<conio.h>

void main()
{
int n,i,res,finres;
char opt ;
clrscr () ;
do
{
res = 0 ;
finres = 0 ;
printf ( " \n Enter the last number " ) ;
scanf("%d",&n);

for ( i = 1 ; i <= n ;i++ )
{
res = ( i * (i+2) * (i+4) ) ;

if ( (i%2) == 0 )
finres = finres - res ;
else
finres = finres + res ;
}
printf ( " \n the result is : %d ", finres) ;
printf ( "Do you want to continue (Y/N)" ) ;
scanf ( "%s",&opt) ;
}while ( (opt == 'y' ) || (opt == 'Y' ) ) ;
}

Answer / anshu porwal

#include<stdio.h>
#include<conio.h>
void main()
{
int x=1,y=3,z=5,n,cal,add=0,add1=0,total,i;
clrscr();
printf(" Enter Any Number to calculate Series for n
terms :");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
cal=(x*y*z);
x++;
y++;
z++;
if(i%2==1)
{
add=add+cal;
}
if(i%2==0)
{
add1=add1+cal;
}
}
total=add-add1;
printf(" Total Of Series Is %d",total);
getch();
}

Answer / guest

#include<iosteram.h>
#include<conio.h>
int seri(int)
{
int r,s;
r=s*m+var;
return r;
}
void main()
{
int a[20],b[20];
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>a[i];
int var=-1;
m=a[i];
cout<<m;
getch();
}

Answer / y hussain reddy

#include<stdio.h>
#include<conio.h>
#include<math.h>

void main()
{
int i,p,n;
long s=0;
puts("nter n\n");
scanf("%d",&n);
for(i=1;i<=n;i++)
{p=pow(-1,i-1);
s+=i*(i+2)*(i+4)*p;
}
printf("ans=%ld",s);
}

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