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)  =  x2k − y2k

Let p(k) be true. Then x2k − y2k is divisible by (x + y)

x2k − y2k  =  λ(x+y)

Step 3 :

put n  =  k + 1

We shall now show that p(k+1) is true. x2(k+1) − y2(k+1) is divisible by (x + y).

 x2(k+1) − y2(k+1)  =   x2k+2 − y2k+2

Add and subtract by x2k y2

 =   x2k+2 x2k y2 x2k y− y2k+2

  =  x2k (x2 - y2) + y2(x2k - y2k)

  =  x2k (x2 - y2) + y2λ

  =  (x + y) (x2k (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 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 + · · · + n2n3/3

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 

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