## INDEX

### Write a c program to find out the sum of series 1^2 + 2^2 + …. + n^2.

Sum of 1^2 + 2^2 + …. + n^2 series in c programming language

#include<stdio.h>

int main(){

int n,i;
int sum=0;

printf("Enter the n i.e. max values of series: ");
scanf("%d",&n);

sum = (n * (n + 1) * (2 * n + 1 )) / 6;

printf("Sum of the series : ");

for(i =1;i<=n;i++){
if (i != n)
printf("%d^2 + ",i);
else
printf("%d^2 = %d ",i,sum);
}

return 0;
}

Sample output:

Enter the n i.e. max values of series: 5
Sum of the series: 1^2 + 2^2 + 3^2 + 4^2 + 5^2 = 55

Mathematical Formula:

Sum of the series 12 + 22 + 32 + … + n2 =
n (n+1) (2n+1)/6

6. Write a c program to find out the sum of given H.P.

preeree said...

thanks...............

Anonymous said...

Help me
Write a C or Java program to find out the sum of series:
1! + 2! + 3! + . . . + N, with N input from keyboard

Md Faiyaz Ansari said...

thank you :)

Anonymous said...

1/x+1/x^2+1/x^3+.....

Ah Lop said...

Very good..

Vibon Keat said...

Dear sir, Can you have me to write total sum of 2^2+10^2+20^2+50^2+100^2 in java??

mohsinahmed alif said...

1^2+3^2+5^2....................nth find out sum use for loop in c

mohsinahmed alif said...

plz........ any one help me.........

monir hossain said...

#include
int main()
{
int n,sum=0,i;
scanf("%d",&n);
for(i=0;i<=n;i++)
{
sum=(sum+i*i);
}
printf("%d",sum);
return 0;
}

eng porag said...

please solutionthis program in C++ to generate the series as 1, 3, 4, 8, 15, 27, 50, ........

eng porag said...

please solution this program in C++ to generate the series as 1, 3, 4, 8, 15, 27, 50, ........

Solaiman Shohag said...

#include
int main()
{
int n,sum=0,i;
scanf("%d",&n);
for(i=1;i<=n;i=i+2)
{
sum=(sum+i*i);
}
printf("%d",sum);
return 0;
}

Tapos Bormon said...

1^2+2^4+3^6+4^8+.............+n^2n for loop c programming code please ans me

Abhra Mitra,1997 said...
This comment has been removed by the author.
Abhra Mitra,1997 said...

your program didn't work. I did a little modification and now it works..

#include
int main()
{
int n,sum=1,i, j=0;
scanf("%d",&n);
for(i=1;i<=n; i++)
{
sum=(sum+(j*j));
j=j+2;
}
printf("%d",sum);
return 0;
}

Cynthia Zemp said...
This comment has been removed by the author.
jonatha ivan sanchez martinez said...

1^2+2^2+n^2 /n^2 en c++ como quedaría me dijeron que fuera dando los pasos

jonatha ivan sanchez martinez said...

que libreria metiste para que funcionara el programa

Tewfiqur Rahman said...