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n2 + 3n is always divisible by which number, provided n is an integer?

n2 + 3n is always divisible by which number, provided n is an integer? Correct Answer 2

P(n) = n2 + 3n P(1) = 1 + 3 P(1) = 4 Let’s assume that P(k) is true and divisible by 4. Therefore, P(k) = k2 + 3k can be written as 4c. We need to check if P(k + 1) is divisible by 4 P(k+1) = (k + 1)2 + 3(k + 1) P(k+1) = k2 + 1 + 2k + 3k + 3 P(k+1) = k2 + 5k + 4 P(k+1) = (k2 + 3k) + 2k + 4 P(k+1) = 4c + 2k + 4 P(k+1) = 4c + 2(k + 2) Clearly the second part of the equation is not divisible by 4. However P(k) = 4c is divisible by 2 and P(k + 1) is also divisible by 2. Therefore, 2 divides P(n).

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