The three Investigators Series

Introduction for three investigators series:

In this article, we will discuss about the three investigators series. The three different series are

1. Arithmetic series

2. Geometric series

3. Power series.

Arithmetic series means that, the sequence of numbers such that the difference between two consecutive members of the sequence is a constant. Geometric series means that, the sequence of numbers such that the ratio between two consecutive members of the sequence is a constant. Power series is a infinite series form. The three investigators series formulas and example problems are given below.

Formulas for three Investigators Series:

Formula for arithmetic sequence:

nth term of the sequence : `a_n = a_1 + (n - 1)d`

Series of the sequence: `S_n` = `(n(a_1 + a_n))/2 `

Formula for geometric sequence:

nth term of the sequence: `a_n = a_1* r^(n-1)`

Series of the sequence: `S_n` = `(a_1(1-r^n))/(1 - r)`

Formula for power series:

`sumn^2 = [n(n+1)(2n+1)]/6`

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Example Problems for three Investigators Series:
Example problem 1:

Find the 13th term of the series 8, 9, 10, 11, 12,....

Solution:

First term of the series, a1 = 8

Difference of two consecutive terms, d = 9 - 8 = 1

n = 13

The formula to find the nth term of an arithmetic series, `a_n = a_1 + (n-1)d`

So, the 16th term of the series 8, 9, 10, 11, 12,... = 8 + (13 - 1) 1

= 8 + 13 * 1

= 8 + 13

After simplify this, we get

= 21

So, the 13th term of the sequence is 21.

Series of the arithmetic sequence: `S_n` = `(n(a_1 + a_n))/2 `

` s_13 = [13(8 + 21)]/(2) `

` s_13 = [13(29)]/(2)`

` s_13 = (377)/(2)`

After simplify this, we get

` s_13 = 188.5`

Example problem 2:

Find out the 6th term of a geometric series if a1 = 80 and the common ratio (C.R) r = 2

Solution:

Here `a_1` = 80, r = 2 and n = 6

Use the formula `a_n = a_1 * r^(n-1)` that gives the nth term to find `a_6` as follows

`a_6 = a_1 * r^(6 -1)`

= 80 * (2)5

= 80 * 32

= 2560.

The 6th term of a geometric sequence is 2560.

Series of the sequence: `S_n` = `(a_1(1-r^n))/(1 - r)`

` s_6 = [80(1 - 2^6)]/(1 - 2)`

`= [80(-64)]/(-1)`

After simplify this, we get

`s_6 = 5120`

Example problem 3:

Find the given series: `1^2 + 3^2 + 5^2 + 7^2 +....... + 22^2`

Solution:

Formula = ` [n(n+1)(2n+1)]/6`

Here n = 22

= ` [22(22+1)(44+1)]/6`

= `[22(23)(45)]/6`

= ` (22770)/6`

After simplify this, we get

= 3795.

The above examples are helpful to study of three investigators series.