Given a quadratic series as given below, the task is to find the sum of the first n terms of this series.
Sn = 3 + 7 + 13 + 21 + 31 + ….. + upto n terms
Examples:
Input: N = 3 Output: 23 Input: N = 4 Output: 44
Approach:
Let the series be represented as
Sn = 3 + 7 + 13 + ... + tn
where
- Sn represents the sum of the series till n terms.
- tn represents the nth term of the series.
Now, to formulate the series, the elements need to be formed by taking the difference of the consecutive elements of the series.
Equation 1: Sn = 3 + 7 + 13 + 21 + 31 +…..+ tn-1 + tn
Equation 2: Sn = 0 + 3 + 7 + 13 + 21 + 31 + …… + tn-1 + tn
(writing the above series by shifting all elements to right by 1 position)
Now, Subtract Equation 2 from Equation 1 i.e. (Equation 1 – Equation 2)
Sn – Sn = (3 – 0) + (7 – 3) + (13 – 7) + (31 – 21) + …… + (tn- tn-1) – tn
=> 0 = 3 + 4 + 6 + 8 + 10 + …… + (tn – tn-1) – tn
In the above series, leaving 3 aside, terms starting from 4 to (tn – tn-1) will form an A.P.
Since the formula of the sum of n terms of A.P. is:
Sn = n*(2*a + (n – 1)*d)/2
which implies,
In series: 4 + 6 + 8 + … + (tn – tn-1)
AP is formed with (n-1) terms.
Hence,
Sum of this series: (n-1)*(2*4 + (n-2)*2)/2
Therefore, the original series:
0 = 3 + (n-1)*(2*4 + (n-2)*2)/2 – tn
where tn = n^2 + n + 1 which is the nth term.
Therefore,
Sum of first n terms of series will be:
tn = n^2 + n + 1
Sn =(n^2) +
Sn = n*(n+1)*(n+2)/6 + n*(n+1)/2 + n
Sn = n*(n^2 + 3*n + 5)/3
(n^2) + Below is the implementation of the above approach:
C++
// C++ program to find sum of first n terms#include <bits/stdc++.h>using namespace std;int calculateSum(int n){ // Sum = n*(n^2 + 3*n + 5)/3 return n * (pow(n, 2) + 3 * n + 5) / 3;}int main(){ // number of terms to be included in the sum int n = 3; // find the Sum cout << "Sum = " << calculateSum(n); return 0;} |
Java
// Java program to find sum of first n termsimport java.util.*;class solution{//function to calculate sum of n terms of the seriesstatic int calculateSum(int n){ // Sum = n*(n^2 + 3*n + 5)/3 return n * (int) (Math.pow(n, 2) + 3 * n + 5 )/ 3;}public static void main(String arr[]){ // number of terms to be included in the sum int n = 3; // find the Sum System.out.println("Sum = " +calculateSum(n));}} |
Python3
# Python 3 program to find sum # of first n termsfrom math import powdef calculateSum(n): # Sum = n*(n^2 + 3*n + 5)/3 return n * (pow(n, 2) + 3 * n + 5) / 3if __name__ == '__main__': # number of terms to be included # in the sum n = 3 # find the Sum print("Sum =", int(calculateSum(n)))# This code is contributed by# Sanjit_Prasad |
C#
// C# program to find sum of first n termsusing System;class gfg{ public double calculateSum(int n) { // Sum = n*(n^2 + 3*n + 5)/3 return (n * (Math.Pow(n, 2) + 3 * n + 5) / 3); }}//driver codeclass geek{ public static int Main() { gfg g = new gfg(); // number of terms to be included in the sum int n = 3; //find the Sum Console.WriteLine( "Sum = {0}", g.calculateSum(n)); return 0; }} |
PHP
<?php// PHP program to find sum// of first n termsfunction calculateSum($n){ // Sum = n*(n^2 + 3*n + 5)/3 return $n * (pow($n, 2) + 3 * $n + 5) / 3;}// Driver Code// number of terms to be // included in the sum$n = 3;// find the Sumecho "Sum = " . calculateSum($n);// This code is contributed by mits?> |
Javascript
<script>// Javascript program to find sum of first n terms// Function to find the quadratic // equation whose roots are a and bfunction calculateSum(n){ // Sum = n*(n^2 + 3*n + 5)/3 return n * (Math.pow(n, 2) + 3 * n + 5 ) / 3;}// Driver Code // Number of terms to be // included in the sumvar n = 3; // Find the Sumdocument.write("Sum = " + calculateSum(n));// This code is contributed by Ankita saini </script> |
Sum = 23
Time Complexity: O(1), since there is no loop or recursion.
Auxiliary Space: O(1), since no extra space has been taken.
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