# 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) : 10k + 3.4 k+2 + 5 is divisible by 9.

10k + 3.4k+2 + 5  = 9m

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

=  10 x 10k + 3.4k+2 .4+ 5

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

=  90m - 30 4k+2 - 50 + 12.4k+2 + 5

=  90m - 18 4k+2 - 45

=  9(10m - 2.4k+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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