Question #92258
Show that 1n3 + 2n + 3n2 is divisible by 2 and 3 for all positive integers n.
1
Expert's answer
2019-08-05T12:58:53-0400

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


1×13+2×1+3×12=1+2+3=61\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×k3+2×k+3×k21\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×(k+1)3+2×(k+1)+3×(k+1)2=1\times (k+1)^3+2\times (k+1) + 3 \times (k+1)^2=


=k3+3×k2+3×k+1+2×k+2+3×k2+6×k+3=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×k3+2×k+3×k2+3×k2+9×k+61\times k^3+2\times k + 3 \times k^2+3\times k^2+9\times k + 6

By assumption, the expression


1×k3+2×k+3×k21\times k^3+2\times k + 3 \times k^2

is divisible by 6. We show that the expression


3×k2+9×k+63\times k^2+9\times k + 6

is also divisible by 6.


3×k2+9×k+6=3(k2+3×k+2)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(k2+3×k+2)=3((2t)2+3×(2t)+2)=3( k^2+3\times k + 2)=3( (2t)^2+3\times (2t) + 2)=

3(4×t2+6×t+2)=6(2×t2+3×t+1)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(k2+3×k+2)=3((2t+1)2+3×(2t+1)+2)=3( k^2+3\times k + 2)=3( (2t+1)^2+3\times (2t+1) + 2)=

=3(4×t2+4×t+1+6×t+3+2)==3(4\times t^2 + 4 \times t + 1 +6 \times t +3 +2)=

=6(2×t2+5×t+3)=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×k2+9×k+63\times k^2+9\times k + 6

is divisible by 6. An expression


1×k3+2×k+3×k2+3×k2+9×k+61\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.



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Comments

Assignment Expert
10.09.20, 20:23

A question does not specify which method is acceptable here. The method of mathematical induction is well-known and it was applied to solve a question.

Dundas mannen
10.09.20, 19:58

No need for Induction

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