MATHEMATICAL INDUCTION FOR DIVISIBILITY

Example 1 :

Using the Mathematical induction, show that for any natural number n, x²ⁿ − y²ⁿ is divisible by x + y.

Solution :

Let p(n) be the statement given by

p(n)  =  x²ⁿ − y²ⁿ   is divisible by (x+y)

Step 1 :

put n = 1

p(1)  =  x²⁽¹⁾ − y²⁽¹⁾

   =  x² - y²

   =  (x + y)(x - y) which is divisible by (x+y)

Hence p(1) is true.

Step 2 :

put n = k

p(k)  =  x^2k − y^2k

Let p(k) be true. Then x^2k − y^2k is divisible by (x + y)

x^2k − y^2k  =  λ(x+y)

Step 3 :

put n  =  k + 1

We shall now show that p(k+1) is true. x^2(k+1) − y^2(k+1) is divisible by (x + y).

 x^2(k+1) − y^2(k+1)  =   x^2k+2 − y^2k+2

Add and subtract by x^2k y²

 =   x^2k+2 - x^2k y² + x^2k y² − y^2k+2

  =  x^2k (x² - y²) + y²(x^2k - y^2k)

  =  x^2k (x² - y²) + y²λ

  =  (x + y) (x^2k (x² - y²) + λy²)

Clearly it is divisible by (x + y)

p(k + 1) is true.

Then p(k) is true  ==>  p(k + 1) is true.

Hence, by the principle of mathematical induction p(k) is true for all n ∈ N x²ⁿ− y²ⁿ is divisible by (x + y) for all n∈N

Example 2 :

By the principle of Mathematical induction, prove that, for n ≥ 1,  1² + 2² + 3² + · · · + n² > n³/3

Solution :

Step 1 :

put n = 1

p(1)  = 1² + 2² + 3² + · · · + 1²  > 1³/3

L.H.S 

Sum of squares  =  n (n+1)(2n+1)/6

  =  1(2)(3)/6  =  1

R.H.S

  =  1/3

L.H.S > R.H.S  

Hence p(1) is true.

Step 2 :

Let us assume that the statement is true for n = k

p(k)  = 1² + 2² + 3² + · · · + k²  > k³/3

We need to show that P(k + 1) is true. Consider,

Step 3 :

Let us assume that the statement is true for n = k + 1

p(k+1) 

1² + 2² + 3² + · · · + (k +1)²  > (k + 1)³/3

1² + 2² + 3² + · · · k²+ (k +1)²  > (k + 1)³/3

1² + 2² + 3² + · · · k²+ (k +1)²  >  (k³/3) + (k + 1)³/3

1² + 2² + 3² + · · · k²+ (k +1)²  >  (1/3) (k³ + 3k² + 6k + 3)

1² + 2² + 3² + · · · k²+ (k +1)² > (1/3) (k³+3k²+3k+1)+(3k+2)

1²+2²+3²+ · · · k²+ (k +1)² > (1/3){(k + 1)³+(3k+2)} > (k+1)³/3

p(k+1) is true.

p(k) is true  ==>  p(k+1) us true.

Hence, by the principle of mathematical induction, p(k) us true for all n ∈ N 

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