Question

Докажите, что при любом натуральном n значение выражения n^3 + 3n^2 + 2n делится нацело на 6. Ответь, пожалуйста, как можно подробнее.

Answer 1

$n^3+3n^2+2n = n(n^2+3n+2) = n(n+1)(n+2)$

Делится ли произведение трех подряд идущих чисел на 6? Конечно да, ведь среди них есть как минимум одно четное число и как минимум одно, которое делится на 3 без остатка.

А если произведение делится на 2 и 3, оно делится на 6

Answer 2

Proof that the expression n^3 + 3n^2 + 2n is divisible by 6 for any natural number n

To prove that the expression n^3 + 3n^2 + 2n is divisible by 6 for any natural number n, we need to show that it leaves no remainder when divided by 6.

Let's break down the expression and analyze it step by step:

1. n^3: This term represents the cube of the natural number n. 2. 3n^2: This term represents three times the square of the natural number n. 3. 2n: This term represents two times the natural number n.

To prove that the expression is divisible by 6, we need to show that each term is divisible by 6.

1. n^3: Any cube of a natural number is divisible by 6. This is because if we consider the remainders when dividing the numbers 1, 2, 3, 4, 5, 6, etc., by 6, we will notice that the remainders repeat in a cycle of length 6: 1, 8, 27, 64, 125, 216, 1, 8, 27, 64, 125, 216, etc. Therefore, n^3 is divisible by 6 for any natural number n.

2. 3n^2: Since n^2 is divisible by 6 (as shown above), multiplying it by 3 will still result in a number divisible by 6.

3. 2n: Any multiple of 2 is divisible by 6 because 6 is a multiple of 2.

Therefore, each term in the expression n^3 + 3n^2 + 2n is divisible by 6, which means that the entire expression is also divisible by 6.

In conclusion, the expression n^3 + 3n^2 + 2n is divisible by 6 for any natural number n.