SUM SERIES

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         SUM OF SERIES


       1.SUM OF SERIES 1+1/2+1/3+1....1/N
      
#include<stdio.h>
main()
{
int i,n;
float sum=0;
printf("Enter value of n\n");
scanf("%d",&n);


for(i=1;i<=n;i++)
{
 sum+=(float)1/i;
}
printf("The value of\n");
for(i=1;i<=n;i++)
{
 if(i<n)
 {
  printf("1/%d+ ",i);
 }
 else
 {
  printf("1/%d= ",i);
 }
}
printf("%f\n",sum);
}
•Output

Enter value of n
8
The value of
1/1+ 1/2+ 1/3+ 1/4+ 1/5+ 1/6+ 1/7+ 1/8= 2.717857

2.SUM OF SERIES 1/1!+1/2!+1/3!+1....+N /N!

main()
{
int i,n;
float sum=0;
printf("Enter value of n\n");
scanf("%d",&n);


for(i=1;i&lt;=n;i++)
{
 sum+=(float)i/factorial(i);
}
printf("The value of\n");
for(i=1;i&lt;=n;i++)
{
 if(i&lt;n)
 {
  printf("%d/%d!+ ",i,i);
 }
 else
 {
  printf("%d/%d!= ",i,i);
 }

}
printf("%f\n",sum);

}

int factorial(int num)
{
 int i,fact=1;
 for(i=1;i&lt;=num;i++)
 {
  fact=fact*i;
 }

 return(fact);
}

•Output 

Enter value of n
5
The value of
1/1!+ 2/2!+ 3/3!+ 4/4!+ 5/5!= 2.708333



      

 3.TO FIND SUM OF SERIES IN A.P

#include<stdio.h>
main()
{
//an=a+(n-1)d
//lastnum=num1+(n-1)diff  this is formula from our variables
//n=((an-a)/d)+1
  int i,n,sum=0,num1,num2,diff,lastnum,series;

  printf("Enter first 2 numbers for finding common Difference\n");
  scanf("%d%d",&num1,&num2);
  printf("Enter last term of series\n");
  scanf("%d",&lastnum);
  diff=num2-num1;
  series=num1;

  n=((lastnum-num1)/diff)+1;
  printf("Number of terms in series is %d\n",n);

  for(i=1;i<=n;i++)
  {
   sum+=series;
   series+=diff;
  }

  series=num1;
  printf("Sum of the series\n");
    for(i=1;i<=n;i++)
  {
     if(i<n)
     {
      printf("%d+",series);
      series+=diff;
     }
     else
     {
      printf("%d=%d\n",series,sum);
      series+=diff;
     }
  }
}
•Output:
Enter first 2 numbers for finding common Difference
1
5
Enter last term of series
53
Number of terms in series is 14
Sum of the series
1+5+9+13+17+21+25+29+33+37+41+45+49+53=378

               4.TO FIND G.P SERIES 

#include<stdio.h>
#include<math.h>
main()
{
//an=a*r^(n-1)
//lastnum=num1*commonratio^(n-1) this is formula from our variables
  int i,n,sum=0,num1,num2,commonratio,series,lastnum;

  printf("Enter first 2 numbers for finding common Difference\n");
  scanf("%d%d",&num1,&num2);
  printf("Enter number of terms in series\n");
  scanf("%d",&n);
  commonratio=num2/num1;
  printf("Common Ratio is %d\n",commonratio);
lastnum=(num1*(pow(commonratio,(n-1))));
  printf("Last term in series is %d\n",lastnum);
 sum=(num1*(1-pow(commonratio,n)))/(1-commonratio);
  series=num1;
  printf("Sum of the series \n");
    for(i=1;i<=n;i++)
  {
     if(i<n)
     {
      printf("%d+",series);
      series*=commonratio;
     }
     else
     {
      printf("%d=%d\n",series,sum);
      series*=commonratio;
     }
  }
}
•Output
Enter first 2 numbers for finding common Difference
10
30
Enter number of terms in series
10
Common Ratio is 3
Last term in series is 196830
Sum of the series
10+30+90+270+810+2430+7290+21870+65610+196830=295240


    5.SUM OF SERIES 1^2+2^2+3^2+......n^2

#include <stdio.h>
#include<math.h>
main()
{
 //1^2+2^2+3^2+......n^2
  int i,num,sum=0;
  printf("Enter number \n");
  scanf("%d",&num);
  for(i=1;i<=num;i++)
  {
   sum+=pow(i,2);
  }
  printf("Sum of series\n");
  for(i=1;i<=num;i++)
  {
   if(i!=num)
   {
      printf("%d^2+ ",i);
    }
    else
    {
     printf("%d^2 = %d\n",i,sum);
    }

  }
}
•Output
Enter number
25
Sum of series
1^2+ 2^2+ 3^2+ 4^2+ 5^2+ 6^2+ 7^2+ 8^2+ 9^2+ 10^2+ 11^2+ 12^2+ 13^2+ 14^2+ 15^2+ 16^2+ 17^2+ 18^2+ 19^2+ 20^2+ 21^2+ 22^2+ 23^2+ 24^2+ 25^2 = 5525

   6.SUM OF SERIES 1^2-2^2+3^2-......+n^2

#include <stdio.h>
 #include<math.h>
 main()
 {
  //1^2-2^2+3^2-......+n^2
   int i,num,sum=0;
   printf("Enter number \n");
   scanf("%d",&num);
   for(i=1;i<=num;i++)
   {
    if(i%2==0)
    {
     sum-=pow(i,2);
     }
     if(i%2==1)
     {
      sum+=pow(i,2);
     }

   }
   printf("Sum of series\n");
   for(i=1;i<=num;i++)
   {
    if(i!=num)
    {
     if(i%2==0)
     {
         printf("%d^2+ ",i);
      }
      else
      {
       printf("%d^2- ",i);
      }

     }
     else
     {
      printf("%d^2 = %d\n",i,sum);
     }

   }
 }
•Output
Enter number
6
Sum of series
1^2- 2^2+ 3^2- 4^2+ 5^2- 6^2 = -22

7.Display series in AP along with sum for given common difference and number of terms

 #include<stdio.h>
main()
{
//an=a+(n-1)d
  int i,n;
  float sum=0,num1,diff,lastnum,seriesFirstNum;

  printf("Enter first number of the series\n");
  scanf("%f",&num1);
  printf("Enter Common difference\n");
  scanf("%f",&diff);
  printf("Enter number of terms in the series\n");
  scanf("%d",&n);
  seriesFirstNum=num1;
lastnum=(float)(num1+((n-1)*diff));
  for(i=1;i<=n;i++)
  {
   sum+=seriesFirstNum;
   seriesFirstNum+=diff;
  }

  seriesFirstNum=num1;
  printf("Terms present in the Series are:\n");
    for(i=1;i<=n;i++)
  {
     if(i<n)
     {
      printf("%.2f,",seriesFirstNum);
      seriesFirstNum+=diff;
     }
     else
     {
      printf("%.2f\nSum of the Series is=%.2f\n",seriesFirstNum,sum);
      seriesFirstNum+=diff;
     }
  }
}
•Output
Enter first number of the series
45
Enter Common difference
6
Enter number of terms in the series
10
Terms present in the Series are:
45.00,51.00,57.00,63.00,69.00,75.00,81.00,87.00,93.00,99.00
Sum of the Series is=720.00

 8.SUM OF SERIES 1^1/1!+2^2/2!+3^3/3!...UPTO N TERMS


#include<stdio.h>
#include<math.h>
int factorial(int);
main()
{
int i,n;
float sum=0;
printf("Enter value of n\n");
scanf("%d",&n);


for(i=1;i<=n;i++)
{
 sum+=(float)(pow(i,i)/factorial(i));
}
printf("The value of\n");
for(i=1;i<=n;i++)
{
 if(i<n)
 {
  printf("%d^%d/%d!+ ",i,i,i);
 }
 else
 {
  printf("%d^%d/%d!= ",i,i,i);
 }

}
printf("%f\n",sum);

}

int factorial(int num)
{
 int i,fact=1;
 for(i=1;i<=num;i++)
 {
  fact=fact*i;
 }

 return(fact);
}
•Output
Enter value of n
5
The value of
1^1/1!+ 2^2/2!+ 3^3/3!+ 4^4/4!+ 5^5/5!= 44.208336

9.SUM SERIES (1+(1+2)+(1+2+3)+(1+2+3+4)+.....till N)

#include<stdio.h>
main()
{
 int i,j,terms,sum=0;
 printf("Enter number oof Terms\n");
 scanf("%d",&terms);
 for(i=1;i<=terms;i++)
 {
  for(j=1;j<=i;j++)
  {
   sum+=j;
  }
 }

 printf("Sum of Series : %d\n",sum);
}
•Output
Enter number oof Terms
6
Sum of Series : 56

10.SUM SERIES (1+(1*2)+(1*2*3)+...till N)

#include<stdio.h>
main()
{
 int i,j,terms,sum=0,mul;
 printf("Enter number oof Terms\n");
 scanf("%d",&terms);
 for(i=1;i<=terms;i++)
 {
  mul=1;
  for(j=1;j<=i;j++)
  {
   mul*=j;
  }
  sum+=mul;
 }

 printf("Sum of Series : %d\n",sum);
}
•Output:
Enter number oof Terms
8
Sum of Series : 46233

11.GENERATE SERIES -1,4,-7,10,-13...till N

#include<stdio.h>
main()
{
 int i,j=1,terms;
 printf("Enter number of terms\n");
 scanf("%d",&terms);
 for(i=1;i<=terms;i++)
 {
  if(i%2==1)
  printf("%d ",(-1*j));
  else
  printf("%d ",j);
  j+=3;
 }
 printf("\n");
}
•Output:
Enter number of terms
10
-1 4 -7 10 -13 16 -19 22 -25 28

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