MATHEMATICAL INDUCTION DIVISIBILITY PROBLEMS

Problem 1 :

Use induction to prove that n³ − 7n + 3, is divisible by 3, for all natural numbers n.

Solution :

Let P(n) = n³ – 7n + 3 is divisible by 3, for all natural numbers n.

Step 1 :

Now  P(l): (l)³ – 7(1) + 3 = -3, which is divisible by 3.

Hence, P(l) is true.

Step 2 :

Let us assume that P(n) is true for some natural number n = k. 

P(k) = K³ – 7k + 3 is divisible by 3

or K³ – 7k + 3 = 3m, m∈ N         (i)

Step 3 :

Now, we have to prove that P(k + 1) is true.

P(k+ 1 )(k + l)³ – 7(k + 1) + 3

  =  k³ + 1 + 3k(k + 1) – 7k— 7 + 3

  =  k³ -7k + 3 + 3k(k + l) - 6

  =  3m + 3[k(k+l)-2]  [Using (i)]

  =  3[m + (k(k + 1) – 2)], which is divisible by 3

Thus, P(k + 1) is true whenever P(k) is true.

So, by the principle of mathematical induction P(n) is true for all natural numbers n.

Problem 2 :

Use induction to prove that 10ⁿ + 3 × 4ⁿ⁺² + 5, is divisible by 9, for all natural numbers n.

Solution :

Step 1 :

n = 1  we have

P(1) ; 10 + 3 ⋅ 64 + 5 = 207 = 9 ⋅ 23

Which is divisible by 9 .

P(1) is true .

Step 2 :

For n =k assume that P(k) is true .

Then P(k) : 10^k + 3.4 ^k+2 + 5 is divisible by 9.

   10^k + 3.4^k+2 + 5  = 9m

10^k = 9m - 3.4^k+2 - 5  ----------(1)

Step 3 :

We have to to prove that P(k+1) is divisible by 9 for

n = k + 1.

P(k + 1) : 10^k+1 + 3.4^k+1+2 + 5

  =  10 x 10^k + 3.4^k+2 .4+ 5 

=  10 ( 9m - 3.4^k+2 - 5 ) + 3.4^k+2 .4+ 5 

=  90m - 30 4^k+2 - 50 + 12.4^k+2 + 5    

=  90m - 18 4^k+2 - 45

=  9(10m - 2.4^k+2 - 5 )

which is divisible by 9 .

 P (k +1 ) is true .

Hence by the Principle of mathematical induction P(n) is true for all n∈N.

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