# C++ Program to Compute Sum of n^2/(n-1) Series

In mathematics, there are many series whose sum is calculated using computer programs to reduces the effort of doing enormous computations. This example program computes the sum of series ( $n^2/(n-1)$). The user has to input the value of n.

To write such programs you must be careful about two things – accuracy and type conversions. Accuracy means correct to at least 3 decimal places because series involves real values. A small difference can produce wrong results.

Sometimes we need to do arithmetic between an integer value and a float value. In such cases, to get the correct value of the expression to convert both operands into the same type using C++ type cast operator.

Before you begin, learn basics of C++ programming. Continue reading if you are familiar with the basics.

Contents

## Problem Definition

We want to write a program that computes the sum of the following series and prints an output to the console. $n^2/(n-1)$

When the program runs, the user provides the value of n and the program computes the sum of all values up to n. The program cannot have a value less than 2 because any value less than 2 is undefined.

For example, $n = 1$ $= 1^2/(1 - 1)$ $= 1/0$ $= undefined$

To understand the problem let us do a manual computation and verify the results using our example program. Suppose $n = 5$ then $= 2^2/(2 - 1)$ + $3^2/(3-1)$ + $4^2/(4-1)$ + $5^2/(5-1)$ $= 4/1 + 9/2 + 16/3 + 25/4$

You can use a calculator or do fraction addition. The final sum of the series is given below. $= 20.08$

Now, you need to verify the output by running the C++ program for the sum of the series given below.

## Program Source Code

#include <cstdlib>
#include <iostream>

using namespace std;

int main()
{

int n,i;
double sum;

//initialize sum to 0

sum = 0.0;

//read value of N

cout << "Enter the value of n:";
cin >> n;

//Compute the value of the sum, but the equation has a restriction.
//The value of n should not be less than 2

for(i=2;i<=n;i++)
{

sum = sum + float(i * i)/float(i-1);

}

//Printing the Output to Console

cout << "\n";
cout << "Sum of the Series n^2/(n-1) =" << " " << sum << endl;
cout << "\n";

system("PAUSE");
return EXIT_SUCCESS;

}

## Output

The output of the program is the same what we received using manual computation method. Therefore, the program works correctly. You may try different inputs.