Question

Prove: The square of a number that is one more than a multiple of 3 is also one more than a multiple of 3

Answer 1

By letting n be a multiple of 3 and expressing the number in question as (n+1) and squaring it, the result is shown to be one more than a multiple of 3, thus proving the original assertion.

Step-by-step explanation:

To prove that the square of a number that is one more than a multiple of 3 is also one more than a multiple of 3, we begin by letting n be a multiple of 3, such that n = 3k, where k is an integer. Then, a number that is one more than n can be written as n+1. The square of this number is (n+1)².

Expanding the square, we have:

(n+1)² = n² + 2n + 1

Substituting n = 3k into the equation gives us:

(3k+1)² = (3k)² + 2×3k + 1

= 9k² + 6k + 1

Notice that 9k² is a multiple of 3 since 9 is a multiple of 3 and k² is an integer. Similarly, 6k is a multiple of 3 since 6 is a multiple of 3 and k is an integer. This makes 9k² + 6k a multiple of 3. Adding 1 to this multiple of 3 gives us an expression that is one more than a multiple of 3.

Therefore, the original assertion is true: the square of a number that is one more than a multiple of 3 is also one more than a multiple of 3.

Answer 2

3(3n² +2n) +1

Step-by-step explanation:

All you have to do is factor the greatest common (integer) factor from the two terms containing n

The greatest common factor of 3·3 = 9 and 3·2 = 6 is 3. That's what goes in the green box outside parentheses. Then the numbers inside parentheses are the remaining factors of those coefficients:

3(3n² +2n) +1

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The fact that you're concerned with multiples of 3 is an additional clue that 3 is a factor of the product. So, the form you're looking for is 3( ) +1.