How to Find the Sum of the Series: 3, 7, 13, 21, 31...

In this article, we will explore how to find the sum of the infinite series 3, 7, 13, 21, 31,...

Identifying the Pattern and Forming the Series

The given series is:

S 3, 7, 13, 21, 31, ...

We can observe that the difference between the consecutive terms is increasing by 4 each time:

7 - 3 4 13 - 7 6 21 - 13 8 31 - 21 10

This indicates that the series is an arithmetic series with the first term a being 3 and the common difference d being 4.

Using the Arithmetic Series Formula

The formula to find the sum of the first n terms of an arithmetic series is given by:

[ S_n frac{n}{2} [2a (n-1)d] ]

Substituting the values for our series:

[ S_n frac{n}{2} [2(3) (n-1)(4)] ]

Simplifying the expression:

[ S_n frac{n}{2} [6 4n - 4] ]

[ S_n frac{n}{2} (4n 2) ]

[ S_n n(2n 1) ]

Concepts and Formulas for Quadratic Sequences

To find the sum of the series 3, 6, 11, ..., we first need to identify the pattern in the series:

3 6 11

The differences between consecutive terms are:

6 - 3 3 11 - 6 5

The second differences are:

5 - 3 2

Since the second differences are constant, the terms can be represented by a quadratic function.

Formulating the Quadratic Equation

We assume the n-th term a_n can be described by a quadratic equation:

[ a_n An^2 Bn C ]

Plugging in the first few terms to find A, B, and C:

For n 1: 12A 1B C 3 → A B C 3 For n 2: 22A 2B C 6 → 4A 2B C 6 For n 3: 32A 3B C 11 → 9A 3B C 11

This gives us a system of equations:

A B C 3 4A 2B C 6 9A 3B C 11

Solving the system:

Subtracting the first equation from the second:

3A B 3

Subtracting the second equation from the third:

5A B 5

Subtracting these two equations gives:

2A 2 → A 1

Substituting A 1 back into equation 1:

3 B 3 → B 0

Substituting A and B into the first equation:

1 0 C 3 → C 2

So, the n-th term is:

[ a_n n^2 2 ]

Sum of the Series Using Quadratic Formulae

The sum Sn of the first n terms is:

[ S_n sum_{k1}^{n} (k^2 2) ]

This can be split into two sums:

[ S_n sum_{k1}^{n} k^2 sum_{k1}^{n} 2 ]

We know:

The formula for the sum of the first n squares is: [ sum_{k1}^{n} k^2 frac{n(n 1)(2n 1)}{6} ] The sum of 2 over n terms is: [ sum_{k1}^{n} 2 2n ]

Therefore:

[ S_n frac{n(n 1)(2n 1)}{6} 2n ]

To combine the terms, we can express 2n with a common denominator:

[ S_n frac{n(n 1)(2n 1)}{6} frac{12n}{6} ]

Simplifying further:

[ S_n frac{n(n 1)(2n 1) 12n}{6} ]

This gives you the sum of the series 3, 6, 11, ... to n terms.