Answer to Question #92258 in Math for Fayoo

Question

Show that 1n3 + 2n + 3n2 is divisible by 2 and 3 for all positive integers n.

Answer 1

Solution. If the number is divisible by 3 and 2, then the number is divisible by 6. (numbers 3 and 2 do not have common divisors). To show that the expression is divisible by 6 for all positive natural numbers, we use the mathematical induction.

Step 1. We show that the expression n³ + 2n + 3n² is divisible by 6 for n = 1

$$1\times 1^3+2\times1+3\times1^2=1+2+3=6$$

The number 6 is divisible by 6.

Step 2. Let the expression n³ + 2n + 3n² is divisible by 6 for n = k.

$$1\times k^3+2\times k + 3 \times k^2$$

We show that the expression 1n3 + 2n + 3n2 is divisible by 6 for n = k+1.

$$1\times (k+1)^3+2\times (k+1) + 3 \times (k+1)^2=$$

$$=k^3+3 \times k^2+3\times k +1+2 \times k +2 +3 \times k^2+6\times k +3$$

We write the expression in the form

$$1\times k^3+2\times k + 3 \times k^2+3\times k^2+9\times k + 6$$

By assumption, the expression

$$1\times k^3+2\times k + 3 \times k^2$$

is divisible by 6. We show that the expression

$$3\times k^2+9\times k + 6$$

is also divisible by 6.

$$3\times k^2+9\times k + 6=3( k^2+3\times k + 2)$$

Obviously, the expressions are divisible by 3. We show that the expression is divisible by 2.

If k is even it can be represented as k = 2t where t is a positive integer. Therefore

$$3( k^2+3\times k + 2)=3( (2t)^2+3\times (2t) + 2)=$$

$$3(4\times t^2+6\times t +2)=6(2\times t^2+3\times t +1)$$

The expression is divisible by 2 for any positive integers t. Consequently, for any even k.

If k is odd it can be represented as k = 2t+1 where t is a positive integer. Therefore

$$3( k^2+3\times k + 2)=3( (2t+1)^2+3\times (2t+1) + 2)=$$

$$=3(4\times t^2 + 4 \times t + 1 +6 \times t +3 +2)=$$

$$=6(2\times t^2 + 5 \times t +3)$$

The expression is divisible by 2 for any positive integers t. Consequently, for any odd k. Therefore, the expression

$$3\times k^2+9\times k + 6$$

is divisible by 6. An expression

$$1\times k^3+2\times k + 3 \times k^2+3\times k^2+9\times k + 6$$

is divisible by 6 as the sum of expressions divisible by 6. Hence the expression n³ + 2n + 3n² is divisible by 6 for n = k+1. According to the principle of mathematical induction, the expression n³ + 2n + 3n² is divisible by 6 for all positive integers n. Therefore, it was shown that the expression n³ + 2n + 3n² is divisible by 3 and 2.