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If (S_n=9n-3n^2), what is the maximum value of the sum of initial terms?

Answer and explanation

Correct answer: (6)

The direct answer is A, 6. The given expression is S_n=9n-3n^2, where n is a positive integer because it counts terms. Calculate successive values: S_1=9(1)-3(1)^2=6; S_2=18-12=6; S_3=27-27=0; S_4=36-48=-12. After this the negative quadratic term makes the values smaller. Another way is to complete the square: S_n=-3n^2+9n=-3(n-3/2)^2+27/4. The real-valued peak is near n=1.5, so among positive integers n=1 and n=2 give the largest integer value, 6. Option A is correct. Option B, 9, is not the value of S_1 or S_2; it is also below neither? Actually the formula gives no 9 for a positive integer n. Option C, 12, is not obtained; S_2 is 6 and S_3 is 0. Option D, 15, is likewise impossible and ignores the negative quadratic term. A common mistake is to use the continuous maximum 27/4 as the answer, but n must be an integer. Memory cue: for a negative quadratic, check nearby integer values around its vertex.

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