# Thread: Help me to solve Five Questions of Discrete Math

1. ## Help me to solve Five Questions of Discrete Math

Five Questions of Discrete Math
help me to solve this

(1)
Show that the sequence a0, a1, a2, … for all integers n ³ 0
an = 3.2n – 2.3n
(2)
Satisfies the recurrence relation
ak = 5ak-1 – 6ak-2 for all integers k ³ 2

(3)
Find the sum of first 100 terms of the given arithmetic sequence -17,-12,-7,-2, 3……

(4)
Find the nth term of the geometric series if
a=125/8 , r=2/5 , n=8

(5)
Whether or not the functions f(x) = x pwer3 and g(x) = x power 1/3 for all x E R
are inverse of one another. Justify your answer.

2. Hello, fahad!

Here are #4 and #5 . . .

(3) Find the sum of first 100 terms of the given arithmetic sequence: -17,-12,-7,-2, 3, ...

The sum of the first n terms is: .S
n .= .½n[2a + (n-1)d]

We have: .a = -17, d = 5, n = 100

Therefore: .S
100 .= .½(100)[2(-17) + 99(5)] .= .23,050

(4) Find the nth term of the geometric series if a = 125/8 , r = 2/5 , n = 8

The nth term is: .a
n .= .a·r^{n-1}

Therefore: .a
8 .= .(125/8)(2/5)^7 .= .16/625

3. Originally Posted by fahad

(1)
Show that the sequence a0, a1, a2, … for all integers n ³ 0
an = 3.2n – 2.3n
(2)
Satisfies the recurrence relation
ak = 5ak-1 – 6ak-2 for all integers k ³ 2
First, I think this is actually one question, secondly, I think it is incomplete. You do not have a full recurrence relation, since you neglected to mention what a_k is for k<2. Anyway, I’d try using induction on this, do you know induction?

(3)
Find the sum of first 100 terms of the given arithmetic sequence -17,-12,-7,-2, 3……
An arithmetic sequence is given by the formula:
a_n = a_1 + (n – 1)d
where a_n is the nth term, a_1 is the first term, n is the current number of the term, and d is the common difference.

We see that:
-17 + 5 = -12
-12 + 5 = -7
-7 + 5 = -2
-2 + 5 = 3
So the common difference is +5
We know the first term is -17, so we have
a_n = -17 + (n – 1)5
or a_n = -22 + 5n, for n = 1,2,3,4,5…

Now the sum of the first n terms of an arithmetic sequence is given by,
S_n = n(a_1 + a_n)/2
We want S_100.

What is the 100th term?
a_100 = -22 + 5(100) = 478

so S_100 = 100(-17 + 478)/2 = 23050

so the sum of the first 100 terms is 23050

(4)
Find the nth term of the geometric series if
a=125/8 , r=2/5 , n=8
The nth term of any geometric SEQUENCE is given by a*r^(n-1) for n = 1,2,3,4 ... Where a is the first term, r is the common ratio, and n is the current number of the term.

A geometric series is a series of the form:
SUM{n=1 to infinity} a*r^(n – 1)
Where a is the first term, r is the common ratio, and n is the current number of the term.

The sum of the first n terms of the series is given by:
S_n = a(1 – r^n)/(1 – r) ...............the formula for the nth term of the SERIES

Since you gave me n = 8, I suppose you are looking for the sum of the first 8 terms.
S_8 = a(1 – r^8)/(1 – r) = (125/8)(1 – (2/5)^8)/(1 – 2/5) = (125/8)(390369/390625)(5/3) = (125/8)(130123/78125) = (1/8)/(130123/625) = 130123/5000 = 26.0246

If you want the 8th term of the SEQUENCE, that would be:

a_8 = (125/8)(2/5)^7 = (125/8)(128/78125) = 16/625 = 0.0256

(5)
Whether or not the functions f(x) = x pwer3 and g(x) = x power 1/3 for all x E R
are inverse of one another. Justify your answer.
Take f(x) = y = x^3, let’s find the inverse of it. How? One way is to switch the x and y and then solve for y. the new function will be the inverse.

y = x^3 ………….switch x and y
For inverse:
x = y^3 …………now solve for y
=> x^(1/3) = y
So the inverse of y is x^(1/3), that is,
f^-1(x) = x^(1/3) = g(x)
so g(x) is the inverse of f(x). You can do this the other way, taking g(x) first and prove f(x) is an inverse of g(x)

another method is to find f(g(x)) and g(f(x)), if the answers to those functions is just x, then they are the inverse of each other.

f(x) = x^3 , g(x) = x^(1/3)
=> f(g(x)) = (x^(1/3))^3 = x
=> g(f(x)) = (x^3)^(1/3) = x
Thus they are the inverses of each other.