When (1-1a)^6 is expanded, the sum of the last three coefficients is:

1962 AMC 12 Problem 12

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12.

When

\left(1-\dfrac1a\right)^6

is expanded, the sum of the last three coefficients is:

22

11

10

-10

-11

The last three terms use powers

a^{-4},

a^{-5},

and

a^{-6}

Include the alternating signs from

(-\frac{1}{a})^k

The last three terms have coefficients

\binom64,\quad-\binom65,\quad\binom66,

so their sum is

15-6+1=10.

Thus, the correct answer is C.