$11 \cdot 6$ When the binomial theorem is used to expand $(a-b)^{n}(n \geqslant 2, a b \neq 0)$, if $a$ $=k b$, (where $k$ is a positive integer), then the sum of the second and third terms is 0, so $n$ equals
(A) $\frac{1}{2} k(k-1)$.
(B) $\frac{1}{2} k(k+1)$.
(C) $2 k-1$.
(D) $2 k$.
(E) $2 k+1$.
(20th American High School Mathematics Examination, 1969)
［Solution］ By the binomial theorem, we have
$$
(a-b)^{n}=a^{n} .-n a^{n-1} b+\frac{n(n-1)}{2} a^{n-2} b^{2}-\cdots+(-1)^{n} b^{n} \text {, }
$$

When $a=k b$, the sum of the second and third terms is
$$
-n(k b)^{n-1} b+\frac{n(n-1)}{2}(k b)^{n-2} b^{2}=0,
$$

which simplifies to $-k+\frac{n-1}{2}=0$, yielding $n=2 k+1$. Therefore, the answer is $(E)$.