Let (1+x+x^2)^10=a_0+a_1x+a_2x^2++a_20x^20.

Question

Let $(1+x+x^2)^{10}=a_0+a_1x+a_2x^2+\cdots+a_{20}x^{20}$. If $(a_1+a_3+a_5+\cdots+a_{19})-11a_2=121k$, then $k$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 239

Accepted Answer

Let $f(x)=(1+x+x^2)^{10}=\displaystyle\sum_{r=0}^{20}a_rx^r$. The sum of odd-indexed coefficients is

$$S_{odd}=a_1+a_3+\cdots+a_{19}=\frac{f(1)-f(-1)}{2}$$

$f(1)=3^{10}$, $f(-1)=(1-1+1)^{10}=1$, so $S_{odd}=\dfrac{3^{10}-1}{2}$.

For $a_2$: write $1+x+x^2=\dfrac{1-x^3}{1-x}$, so $f(x)=\dfrac{(1-x^3)^{10}}{(1-x)^{10}}=\left(\displaystyle\sum_{k=0}^{10}(-1)^k\binom{10}{k}x^{3k}\right)\left(\displaystyle\sum_{r=0}^{\infty}\binom{r+9}{9}x^r\right)$

The coefficient of $x^2$ comes only from $k=0$ (since $3k\ge3$ for $k\ge1$) times the $x^2$ term of the second series: $a_2=1\cdot\binom{11}{9}=55$.

So $$S_{odd}-11a_2 = \frac{3^{10}-1}{2}-11\times55 = \frac{59049-1}{2}-605 = 29524-605=28919$$

$$121k=28919\quad\Rightarrow\quad k=\frac{28919}{121}=239$$

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