# Sum of first N terms of Quadratic Sequence 3 + 7 + 13 + …

• Difficulty Level : Basic
• Last Updated : 08 Apr, 2021

Given a quadratic series as given below, the task is to find sum of first n terms of this series.

Sn = 3 + 7 + 13 + 21 + 31 + ….. + upto n terms

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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 above series, leaving 3 aside, terms starting from 4 to (tn – tn-1) will form an A.P.
Since formula of 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) + n + (1)
Sn = n*(n+1)*(n+2)/6 + n*(n+1)/2 + n
Sn = n*(n^2 + 3*n + 5)/3

Below is the implementation of above approach:

## C++

 // C++ program to find sum of first n terms #include 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 pow def calculateSum(n):         # Sum = n*(n^2 + 3*n + 5)/3    return n * (pow(n, 2) + 3 * n + 5) / 3 if __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

 

## Javascript

 
Output:
Sum = 23

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