program to count number of digits present in a number.

//program to count number of digits present in a number.

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

void main()
{
   int n,rem,count=0;
   printf("enter a number for 'n'\n");
  scanf("%d",&n);

  while(n!=0)
    {

        rem=n%10;

       count++;

        n=n/10;    

   }

printf("total digits =%d",count);

getch();

}

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logics in mind:-

>first we enter a number(for 'n').

->If we have number 123 then it has 3 digits and they are 3, 2 and 1 or 1,2 and 3. It means, first we have to get 3 then 2 and then 1 and we have to count them.        

                  ->For this, 

                 ->we have to divide by 10 to get last digit as remainder 

                 ->we also use 'count' to start counting  digits.

                -> to get second digit, we get first 12 and for this , we use 123/10.It is done to get integer                           part  only.

->We repeat this until the value reaches 0. We use loop for this as shown above in the code.     

->at last we display total number of digits stored in variable 'count'.   

program to get sum of given series with................nth terms.

//program to get sum of x+ x²/2+x³/3+x⁴/4+.............................nth terms.
#include<stdio.h>

#include<conio.h>

#include<math.h>

void main()
{
  float i,n,x,m=1,sum=0;

   printf("enter  positive number for 'n'\n");
  scanf("%f",&n);

printf("enter value for 'x'\n");

scanf("%f",&x);

  for(i=1;i<=n;i++)
    {               

          sum=sum+(pow(x,m)/i);

           m=m+1;

     }

  printf("the sum=%f",sum);

getch();
}

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logics in mind:
->enter a number for range for which you want to get sum, say 'n'.

->let a variable 'sum' with initial value '0'

->here we also need value for 'x' so let's input that

->we get sum using formula sum=sum+(pow(x,m)/i); because in terms x+ x²/2+x³/3+x⁴/4+..............  ,

    ->we have two components namely numerator and denominator. Let's look at numerator;they are

           x¹,x²,x³,x⁴.....it can be formulated as pow(x,m) .  .here the power (m) goes on increasing.
   ->similarly for denominator, it is simply 1,2,3,4.... so it can be used from loop (variable i).

->we find sum inside loop

->at last we display the final sum.

   here 'pow' means finding power. it exists inside math.h

program to get sum of 1+3/4+7/16+15/64..............................nth terms.

//program to get sum of 1+3/4+7/16+15/64..............................nth terms.
#include<stdio.h>

#include<conio.h>

void main()
{
   float i,n,k=2,sum=0,m=0;

   printf("enter  positive number for 'n'\n");
  scanf("%f",&n);

  for(i=1;i<=n;i++)
    {               

          sum=sum+(pow(k,i)-1)/(pow(k,m));

          m=m+2;    

     }

  printf("the sum=%f",sum);

getch();
}

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logics in mind:
->enter a number for range for which you want to get sum, say 'n'.

->let a variable 'sum' with initial value '0'

->we get sum using formula sum=sum+(pow(k,i)-1)/(pow(k,m)) because in terms 1+3/4+7/16+15/64...    ,

    ->we have two components namely numerator and denominator. Let's look at numerator;they are

           1,3,7,16.....it can be formulated as pow(k,i)-1 that is    2¹-1  .here the power goes on increasing
   ->similarly for denominator, it has power of 2 i.e.
                 2⁰,2²,2^{4. we can see here that power is increasing by 2.}

                for this we have,pow(k,m) then m=m+2
->we find sum inside loop

->at last we display the final sum.

program to get sum of given series with ..................nth terms.

//program to get sum of 1²x1+2³x3+3⁴x5+..............................nth terms.
#include<stdio.h>

#include<conio.h>

void main()
{
   float i,n,k=2,m=1sum=0;

   printf("enter  positive number for 'n'\n");
  scanf("%f",&n);

  for(i=1;i<=n;i++)
    {               

          sum=sum+(pow(i,k)*m);

           k=k+1;

           m=m+2;

     }

  printf("the sum=%f",sum);

getch();
}

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logics in mind:
->enter a number for range for which you want to get sum, say 'n'.

->let a variable 'sum' with initial value '0'

->we get sum using formula sum=sum+(pow(i,k)*m) because in terms 1²x1+2³x3+3⁴x5+..     ,

        the first value is i(1),second is k(2) and third is m(1).the values in second and all other terms are changing 

   i----by 1
   k by 1
  m by 2     
  so

        we put these values  inside loop.

->at last we display final sum.

program to get sum of (2x3)/5+(4x5)/7+(6x7)/9+,.... to nth term.

//program to get sum of (2x3)/5+(4x5)/7+(6x7)/9+,....           to nth term.
#include<stdio.h>

#include<conio.h>

void main()
{
   float i,n,k=2,m=3,p=5,sum=0;

      printf("enter  positive number for 'n'\n");
  scanf("%f",&n);

  for(i=1;i<=n;i++)
    {               

          sum=sum+(k*m)/p;

           k=k+2;

           m=m+2;

          p=p+2;        }

  printf("the sum=%f",sum);

getch();
}

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logics in mind:
->enter a number for range for which you want to get sum, say 'n'.

->let a variable 'sum' with initial value '0'

->we get sum using formula sum=sum+(k*m)/p because in terms (2x3)/5+(4x5)/7+(6x7)/9+,...     ,

        the first value is k(2),second is m(3) and third is p(5).the values in second and all other terms are changing 

       by +2 so

        we put

          k=k+2;

           m=m+2;

          p=p+2; inside loop.

->at last we display final sum.