<P>Answer

2

Step-by-step explanation:

Consider an initial value for which the statement is true. It is to be shown that the statement is true for N = initial value.

Step 2 − Assume the statement is true for any value of n = k. Then prove the statement is true for n = k+1. We actually break n = k+1 into two parts, ONE PART is n = k (which is already proved) and try to prove the other part.

Problem 1

3n−1 is a multiple of 2 for n = 1, 2, ...

Solution

Step 1 − For n=1,31−1=3−1=2 which is a multiple of 2

Step 2 − Let us assume 3n−1 is true for n=k, Hence, 3k−1 is true (It is an assumption)

We have to prove that 3k+1−1 is also a multiple of 2

3k+1−1=3×3k−1=(2×3k)+(3k−1)

The first part (2×3k) is certain to be a multiple of 2 and the second part (3k−1) is also true as our previous assumption.

Hence, 3k+1–1 is a multiple of 2.

So, it is proved that 3n–1 is a multiple of 2.

Problem 2

1+3+5+...+(2n−1)=n2 for n=1,2,…

Solution

Step 1 − For n=1,1=12, Hence, step 1 is satisfied.

Step 2 − Let us assume the statement is true for n=k.

Hence, 1+3+5+⋯+(2k−1)=k2 is true (It is an assumption)

We have to prove that 1+3+5+...+(2(k+1)−1)=(k+1)2 also holds

1+3+5+⋯+(2(k+1)−1)

=1+3+5+⋯+(2k+2−1)

=1+3+5+⋯+(2k+1)

=1+3+5+⋯+(2k−1)+(2k+1)

=k2+(2k+1)

=(k+1)2

So, 1+3+5+⋯+(2(k+1)−1)=(k+1)2 hold which satisfies the step 2.

Hence, 1+3+5+⋯+(2n−1)=n2 is proved.

Problem 3

Prove that (ab)n=anbn is true for every natural number n

Solution

Step 1 − For n=1,(ab)1=a1b1=ab, Hence, step 1 is satisfied.

Step 2 − Let us assume the statement is true for n=k, Hence, (ab)k=akbk is true (It is an assumption).

We have to prove that (ab)k+1=ak+1bk+1 also hold

Given, (ab)k=akbk

Or, (ab)k(ab)=(akbk)(ab) [Multiplying both side by 'ab']

Or, (ab)k+1=(aak)(bbk)

Or, (ab)k+1=(ak+1bk+1)

Hence, step 2 is proved.

So, (ab)n=anbn is true for every natural number n.

Strong Induction

Strong Induction is another form of mathematical induction. Through this induction technique, we can prove that a propositional function, P(n) is true for all positive INTEGERS, n, using the following steps −

Step 1(Base step) − It proves that the initial proposition P(1) true.

Step 2(Inductive step) − It proves that the conditional statement [P(1)∧P(2)∧P(3)∧⋯∧P(k)]→P(k+1) is true for positive integers k.