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**Principle of Mathematical Induction**

Let P(n) be a mathematical statement, where n is a natural number, such that:

(i) The statement is true for n = 1, or P(1) is true.

(ii) If the statement is true for n = k, where k is a positive integer, then the statement is also true for n = k + 1.

P(k) is true â‡’ P(k + 1) is true.

Then, P(n) is true for all natural numbers n.

Principle one is a statement of fact, while principle two is a condition.

If P(n) is true for all n â‰¥ 2, then step 1 starts from n = 2 and we verify the result of P(2).

Next, if the second principle is true for n = k, then it is also true for n = k + 1.

Using the principle of mathematical induction, prove the following statements:

(i) 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2n-1)(2n+1) = n/(2n + 1)

(ii) For all n âˆˆ N; (2^{3n} - 1) is divisible by 7.

(i) 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2n-1)(2n+1) = n/(2n + 1)

P(n): 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2n-1)(2n+1) = n/(2n + 1)

For n = 1,

P(1): 1/1.3 = 1/((2x1)+1) â‡’ 1/3 = 1/3

âˆ´ P(1) is true.

Assume that P(n) is true for some k âˆˆ N.

P(k): 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2k-1)(2k+1) = k/(2k + 1) ----------- (1)

To prove that P(k + 1) is true, add the next term of the series to both sides of the statement.

1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2k-1)(2k+1) + 1/([2(k+1)-1][2(k+1)+1])= k/(2k + 1) + 1/([2(k+1)-1][2(k+1)+1])

= k/(2k + 1) + 1/((2k + 1)(2k + 3))

= (k/(2k + 3) + 1)/((2k+1)(2k+3))

=(2k^{2}+3k+1)/((2k+1)(2k+3))

=((2k+1)(k+1))/((2k+1)(2k+3))

=(k+1)/(2k+3)

=(k+1)/(2(k+1)+1)

=P(k+1)

âˆ´ P(k+1) is true.

Hence, P(n) is true for all natural numbers n.

We have to prove that the expression is divisible by seven for all natural numbers n.

P(n): (2^{3n} - 1) is divisible by 7.

For n = 1, P(1): 2^{3x1} - 1 = 7, which is divisible by 7.

âˆ´ P(1) is true.

Let P(n) be true for some k âˆˆ N.

âˆ´ P(k):(2^{3k} -1) is divisible by 7 ------- (1)

Need to prove that P(k + 1) is true.

P(k + 1): 2^{3(k+1)} - 1

= 2^{3k+3} - 1

=2^{3k}.2^{3} - 1

=2^{3k}.8 - 1

=2^{3k}.(7+1) - 1

=2^{3k}.7 + (2^{3k} - 1)

2^{3k}.7is a multiple of 7.

2^{3k}-1 is divisible by 7 from equation 1.

âˆ´ P(k+1) is divisible by 7.

Hence, (2^{3n} - 1) is divisible by 7 for all natural numbers n.

**Principle of Mathematical Induction**

Let P(n) be a mathematical statement, where n is a natural number, such that:

(i) The statement is true for n = 1, or P(1) is true.

(ii) If the statement is true for n = k, where k is a positive integer, then the statement is also true for n = k + 1.

P(k) is true â‡’ P(k + 1) is true.

Then, P(n) is true for all natural numbers n.

Principle one is a statement of fact, while principle two is a condition.

If P(n) is true for all n â‰¥ 2, then step 1 starts from n = 2 and we verify the result of P(2).

Next, if the second principle is true for n = k, then it is also true for n = k + 1.

Using the principle of mathematical induction, prove the following statements:

(i) 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2n-1)(2n+1) = n/(2n + 1)

(ii) For all n âˆˆ N; (2^{3n} - 1) is divisible by 7.

(i) 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2n-1)(2n+1) = n/(2n + 1)

P(n): 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2n-1)(2n+1) = n/(2n + 1)

For n = 1,

P(1): 1/1.3 = 1/((2x1)+1) â‡’ 1/3 = 1/3

âˆ´ P(1) is true.

Assume that P(n) is true for some k âˆˆ N.

P(k): 1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2k-1)(2k+1) = k/(2k + 1) ----------- (1)

To prove that P(k + 1) is true, add the next term of the series to both sides of the statement.

1/1.3 + 1/3.5 + 1/5.7 + .......+ 1/(2k-1)(2k+1) + 1/([2(k+1)-1][2(k+1)+1])= k/(2k + 1) + 1/([2(k+1)-1][2(k+1)+1])

= k/(2k + 1) + 1/((2k + 1)(2k + 3))

= (k/(2k + 3) + 1)/((2k+1)(2k+3))

=(2k^{2}+3k+1)/((2k+1)(2k+3))

=((2k+1)(k+1))/((2k+1)(2k+3))

=(k+1)/(2k+3)

=(k+1)/(2(k+1)+1)

=P(k+1)

âˆ´ P(k+1) is true.

Hence, P(n) is true for all natural numbers n.

We have to prove that the expression is divisible by seven for all natural numbers n.

P(n): (2^{3n} - 1) is divisible by 7.

For n = 1, P(1): 2^{3x1} - 1 = 7, which is divisible by 7.

âˆ´ P(1) is true.

Let P(n) be true for some k âˆˆ N.

âˆ´ P(k):(2^{3k} -1) is divisible by 7 ------- (1)

Need to prove that P(k + 1) is true.

P(k + 1): 2^{3(k+1)} - 1

= 2^{3k+3} - 1

=2^{3k}.2^{3} - 1

=2^{3k}.8 - 1

=2^{3k}.(7+1) - 1

=2^{3k}.7 + (2^{3k} - 1)

2^{3k}.7is a multiple of 7.

2^{3k}-1 is divisible by 7 from equation 1.

âˆ´ P(k+1) is divisible by 7.

Hence, (2^{3n} - 1) is divisible by 7 for all natural numbers n.