The code runs well but I'm a beginner and I want to make it more user friendly just out of curiosity.... How to make it more user friendly?

Question

#include<stdio.h>

int main(){

    int i, n;
    long sum=0;
    
    puts("Enter the number: ");
    scanf("%d", &n);


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

        printf("(%d * %d) = %d\n", i, i, (i*i) );

    }

    printf("\nThe sum of the series is %ld\n", sum);

    return 0;
}

suppose I enter the input as 5, then the operation goes like this 1 * 1 = 1, 2 * 2 = 4, 3 * 3 = 9, 4 * 4 = 16, 5 * 5 = 25..... and then, I want to show the sum of every individual product... I want the output to be like this:

(1 * 1) = 1

(2 * 2) = 4

(3 * 3) = 9

(4 * 4) = 16

(5 * 5) = 25

The sum of the series is 1 + 4 + 9 + 16 + 25 = 55

Answer 1

There is a few things you could do just as a suggestion for beginner legibility

sum = sum + (i * i); 

i*i can be pulled out to

int iSquared = i * i;

then store

int nextProduct = sum + iSquared;

and finally simplify the summation with

sum += nextProduct

Bringing it all together you'd have

#include<stdio.h>

int main(){

    int i, n;
    long sum=0;
    
    puts("Enter the number: ");
    scanf("%d", &n);


    for(i=1; i<=n; i++){
        int iSquared = i * i; 
  
        int nextProduct = sum + iSquared;

        sum += nextProduct;

        printf("(%d * %d) = %d\n", i, i, iSquared);

    }

    printf("\nThe sum of the series is %ld\n", sum);

    return 0;
}

It's highly debatable if this is more user friendly honestly, but I still think it's more clear to have each step broken out logically especially when first trying to learn C++ or programming in general.

Answer 2

Here's my attempt at your requirements.

I saved every product in its own array during the loop, then print them one by one again in the final line.

#include <stdio.h>
#include <stdlib.h>

int main(void) {
    unsigned i, n; // no negative numbers used
    long unsigned sum = 0;
    long unsigned parcel[2000]; // could use VLA or dynamic memory

    printf("Enter the number: ");      // write the input on the
    fflush(stdout);                    // same line as the prompt

    if (scanf("%u", &n) != 1) {
        fprintf(stderr, "Input error.\n");
        exit(EXIT_FAILURE);
    }
    if (n > 2000) {
        fprintf(stderr, "Sorry, max number is 2000.\n");
        exit(EXIT_FAILURE);
    }

    for (i = 0; i < n; i++) { // I like to start for loops at 0
        long unsigned product = (i + 1) * (i + 1);
        sum = sum + product;
        printf("(%u * %u) = %lu\n", i + 1, i + 1, product);
        // save product for final output
        parcel[i] = product;
    }

    printf("\nThe sum of the series is %lu", parcel[0]);
    // exception for starting for loops at 0: the 0th element was printed above
    for (i = 1; i < n; i++) printf(" + %lu", parcel[i]);
    printf(" = %lu\n", sum);
    return 0;
}

Answer 3

For a start, you can do it in the loop itself:

printf("\nThe sum of the series is ");
for (i = 1; i <= n; i++){
    sum = sum + (i * i);
    
    // Don't print a leading + for the first number to avoid
    // "The sum of the series is + 1 + 4 + 9 + 16 + 25 = 55"
    if (i > 1) {
        printf(" + ");
    }

    // Print the current term
    printf("%d", i * i);
}
printf(" = %ld\n", sum);

This outputs, for

• n = 2:

The sum of the series is 1 + 4 = 5

• n = 5:

The sum of the series is 1 + 4 + 9 + 16 + 25 = 55

But it also outputs, for

• n = 0:

The sum of the series is  = 0

• n = 1:

The sum of the series is 1 = 1

Handling user input is up to you.

---

If you also wants to output the middle terms, you can iterate over the terms twice:

for (int i = 1; i <= n; i++){
    sum = sum + (i * i);
    printf("(%d * %d) = %d\n", i, i, i*i);
}
    
printf("The sum of the series is ");
for (int i = 1; i <= n; i++){
    if (i > 1) printf(" + ");
    printf("%d", i*i);
}
printf(" = %ld.\n", sum);

Outputs

(1 * 1) = 1
(2 * 2) = 4
(3 * 3) = 9
(4 * 4) = 16
(5 * 5) = 25
The sum of the series is 1 + 4 + 9 + 16 + 25 = 55.