**Table of Contents**

(i) Showing the result of S_{2} and S_{3} for the A.P series. 3

(ii) Calculating the new function by adobe result. 5

Answer:

The sequence of the numbers is,

2, 10, 18, 26, ……, (8n - 6)

This is the sequence of the Arithmetic Progression (A.P) of the first n numbers of values of this series.

The first term of this A.P series is 2 and the last term of this A.P series is (8n - 6).

The sum of this A.P series is given by the form of,

S_{n} = n/2 (8n - 4)

Here,

n = numbers of terms

S_{n} = Sum of the n numbers A.P series

The given inputs by the questions are,

The first term is denoted = a

= 2

The last term is denoted = l

= 8n - 4

Common difference of the two terms is,

d = (value of any term - value of the term in before)

d = 10 - 2

d = 8

The first two terms of this A.P series are 2 and 10

By normally, the sum of the first two terms is = 2 + 10

= 12

Now calculating the outcomes of the sum of first two terms as S_{2} according to this A.P series……..

Here,

n = 2 as the numbers of terms is 2.

S_{n} = n/2 (8n -4)

S_{2} = 2/2 × {(8 × 2) - 4} [after putting the value of n in the equation]

S_{2} = 1 × {16 - 4}

S_{2} = 1 × 12

S_{2} = 12

So it is clear that the result by using the form of S_{n} = n/2 (8n -4) for calculating the sum of first two terms is equal to the normal result. (Proved)

Again, the first three terms of this A.P series are 2, 10 and 18

By normally, the sum of the first two terms is = 2 + 10 + 18

= 30

Again, calculating the outcomes of the sum of first three terms S_{3} according to this A.P series…….

Here n = 3 as the numbers of terms is 3

S_{n} = n/2 (8n -4)

S_{3} = 3/2 × {(8 × 3) - 4} [after putting the value of n in the equation]

S_{3} = 3/2 × {24 - 4}

S_{3} = 3/2 × 20

S_{3} = 60/2

S_{3} = 30

So it is clear that the result by using the form of S_{n} = n/2 (8n -4) for calculating the sum of first three terms is equal to the normal result.

(Proved)

So the two results of S_{2} and S_{3} clearly indicates that the function of the sum of this A.P series is based on this S_{n} = n/2 (8n - 4) function by which any sum of these series is easily calculated.

Now calculating by using the above result……...

_{n=1}Σ^{15} (8n -6)

= (1st term + 2nd term + 3rd term + ……. + last term)

= n/2 (8n - 4)

= 15/2 × {(8 × 15) - 4} [after putting the value of n in the equation]

= 15/2 × {120 - 4}

= 15/2 × 116

= 870

So the result is 870.

Here,

In the A.P series of 2, 10, 18, ….., (8n - 6)

The first term is a = 2

The last term is = (8n - 6)

The common difference is d = value of term - value of previous term

= 10 - 2

= 8

Using the way of the Mathematical Induction we can get that,

S_{n} = 2 + 10 + 18 + 26 + …. + (8n - 6)

After using the formula of summation we get,

S_{n} = n/2 × [2a + (n - 1) × d]

S_{n} = n/2 × [2 × 2 + (n - 1) × 8]

S_{n} = n/2 × [4 + 8n - 8]

S_{n} = n/2 × [8n - 4]

S_{n} = n/2 × 8n - n/2 × 4

S_{n} = 8n^{2}/2 - 4n/2

S_{n} = 4n^{2} - 2n

Then it is concluded that 2 + 10 + 18 + 26 + …. + (8n - 6) = 4n^{2} - 2n by the above result.

So it is proved.

Answer:

F: R → R and F^{-1} : R → R are two functions

The first function is,

F(x) = 3x + 2 where x ∈ R

In other cases the second function is,

F^{-1}(y) = (y - 2)/3 where y ∈ R

Now in first step,

Considering x as a function of y like x = F^{-1}(y)

So,

F(x) = 3x + 2 where x ∈ R

F {f^{-1}(y)} = 3 × (y - 2)/3 + 2 where y ∈ R

F {f^{-1}(y)} = y - 2 + 2

F {f^{-1}(y)} = y

Now in the second step,

Considering y as a function of x like y = F(x)

So,

F^{-1}(y) = (y - 2)/3 where y ∈ R

F^{-1}{f(x)} = {(3x + 2) - 2}/3 where x ∈ R

F^{-1}{f(x)} = 3x/3

F^{-1}{f(x)} = x

Measuring the result of adobe two steps,

The function of x shows the result of y and the other hand the function of y shows the result of x. This result is possible only one time when these functions come is the case of inverse function.

Therefore it can be concluded that these two functions are inverse to each other. (Proved)