1969 AMC 12 Problem 16

Attempt Problem 16 of the 1969 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1969 AMC 12 solutions, or check the answer key.

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16.

When

(a-b)^n,

n\ge2,

ab\ne0,

is expanded by the binomial theorem, it is found that, when

a=kb,

where

k

is a positive integer, the sum of the second and third terms is zero. Then

n

equals:

\dfrac12k(k-1)

\dfrac12k(k+1)

2k-1

2k

2k+1

The second and third terms are

-na^{n-1}b

and

\binom n2a^{n-2}b^2

Substitute

a=kb

and cancel the common nonzero factor

The sum of the second and third terms is

-na^{n-1}b+\frac{n(n-1)}2a^{n-2}b^2.

Substituting

a=kb

and dividing by the nonzero quantity

na^{n-2}b

gives

-k+\frac{n-1}{2}=0.

Therefore

n=2k+1.

Therefore, the correct answer is E.

Problem 16 in other years