- Related Questions & Answers
- Program to find sum of series 1*2*3 + 2*3*4+ 3*4*5 + . . . + n*(n+1)*(n+2) in C++
- C++ program to find the sum of the series 1 + 1/2^2 + 1/3^3 + …..+ 1/n^n
- Program to find sum of series 1 + 2 + 2 + 3 + 3 + 3 + .. + n in C++
- C++ program to find the sum of the series (1*1) + (2*2) + (3*3) + (4*4) + (5*5) + … + (n*n)
- Java Program to Find sum of Series with n-th term as n^2 – (n-1)^2
- C++ Program to find the sum of a Series 1/1! + 2/2! + 3/3! + 4/4! + …… n/n!
- Python Program to find the sum of a Series 1/1! + 2/2! + 3/3! + 4/4! +…….+ n/n!
- C++ program to find the sum of the series 1/1! + 2/2! + 3/3! + 4/4! +…….+ n/n!
- Python Program for Find sum of Series with the n-th term as n^2 – (n-1)^2
- Program to find sum of 1 + x/2! + x^2/3! +…+x^n/(n+1)! in C++
- Program to find sum of series 1 + 1/2 + 1/3 + 1/4 + .. + 1/n in C++
- C++ program to find the sum of the series (1/a + 2/a^2 + 3/a^3 + … + n/a^n)
- C/C++ Program to Find the sum of Series with the n-th term as n^2 – (n-1)^2
- Program to find Sum of a Series a^1/1! + a^2/2! + a^3/3! + a^4/4! +…….+ a^n/n! in C++
- Maximum value of |arr[0] – arr[1] - + |arr[1] – arr[2] - + … +|arr[n – 2] – arr[n – 1] - when elements are from 1 to n in C++

- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who

A series is a sequence of numbers that have some common traits that each number follows. These mathematical series are defined based on some mathematical logic like every number increases by the same interval( arithmetic progression), every number is increased by the same multiple( geometric progression), and many other patterns.

To find the sum of a series we need to evaluate the series and make a general formula for it. But in the series that is no common declaration that takes place so we have to go through the classical approach by adding each number of the series to a sum variable.

Let's take an example that will make the logic more clear,

sum of series upto 7

sum(7) = 1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2} + 7^{2} = 455

#include <stdio.h> int main() { int i, n, sum=0; n=17 ; for ( i = 1; i <= n; i++) { sum = sum + (2 * i - 1) * (2 * i - 1); } printf("The sum of series upto %d is %d", n, sum); }

The sum of series upto 17 is 6545

Advertisements