MTH202

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Maximum Marks: 12
Due Date: June 02, 2009



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Question 1; Mark: 7
Use mathematical induction to prove that for all integers n≥1,
is divisible by 3.

5^n-2^n

is divisible by 3.

P(1) is true
For n = 1, L.H.S
of P(5^1-2^1)
= 5-2
= 3
Hence the equation is true for 5^n-2^n is divisible by 3
Base Case: Let n = 0. Then
15^n - 1 = 15^0 - 1 = 1 - 1 = 0, and 0 is obviously divisible by 7.
Let n = 1. Then
15^1 - 1 = 15 - 1 = 14, and 14 is obviously divisible by 7.

Therefore, the formula holds true for n = 0 and n = 1.

Induction Hypothesis: Assume the formula holds true for up to n = k for some value k. i.e. assume that

15^k - 1 is divisible by 7.

(We want to prove that the formula holds true for n = k + 1, i.e. that 15^(k + 1) - 1 is divisible by 7)

BUT, what exactly is 15^(k + 1) - 1 ? Let's evaluate that.

15^(k + 1) - 1 = 15*(15^k) - 1

What I'm going to do is split 15*(15^k) as 14(15^k) + 1(15^k).

= 14(15^k) + 1(15^k) - 1
= 14(15^k) + (15^k) - 1

I'm going to place the last two terms in square brackets to prove a point.

= 14(15^k) + [ (15^k) - 1 ]

Look at this expression; the second half of this expression is divisible by 7 because 15^k - 1 being divisible by 7 is our induction hypothesis.
The first half, 14(15^k), is OBVIOUSLY divisible by 7, since
14(15^k) = (7*2)(15^k).

That means we have the sum of two things divisible by 7, which means as a whole,

14(15^k) + [ (15^k) - 1 ]

is divisible by 7, and the above expression is equal to

15^(k + 1) - 1

Therefore, the formula holds true for n = k + 1.

And thus, by the Principle of Mathematical Induction,

15^n - 1 is divisible by 7 for all integers n >= 0.


Question 2; Mark: 5
A club consists of four memebers.How many sample points are in the sample space when three officers; president, secretary and treasurer, are to be chosen?
Kindly wait for the second questions solution