n is a positive integer. Explain whether, for all n, (7n+2)2−(7n−1)2(7n+2)^{2}-(7n-1)^{2} is a multiple of 3. Answer ___

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Worked solution

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1. Recognise and apply the difference of two squares identity.

Let a=7n+2a=7n+2 and b=7n−1b=7n-1.

(7n+2)2−(7n−1)2=[(7n+2)+(7n−1)][(7n+2)−(7n−1)](7n+2)^2-(7n-1)^2=[(7n+2)+(7n-1)][(7n+2)-(7n-1)]

2. Simplify the two factors.

(7n+2)+(7n−1)=14n+1(7n+2)+(7n-1)=14n+1
(7n+2)−(7n−1)=7n+2−7n+1=3(7n+2)-(7n-1)=7n+2-7n+1=3

So
(7n+2)2−(7n−1)2=(14n+1)(3).(7n+2)^2-(7n-1)^2=(14n+1)(3).

3. Conclude whether the expression is always a multiple of 3.

(14n+1)(3)=3(14n+1).(14n+1)(3)=3(14n+1).

Since nn is a positive integer, 14n+114n+1 is an integer. Hence 3(14n+1)3(14n+1) is a multiple of 3.

Answer

Yes. Since (7n+2)2−(7n−1)2=3(14n+1)(7n+2)^2-(7n-1)^2=3(14n+1), it is a multiple of 3 for all positive integers nn.

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