Let

Find the sum of odd-indexed coefficients using $P(1)$ and $P(-1)$.

Let $S_{\text{odd}}=a_1+a_3+a_5+\cdots+a_{19},$ $S_{\text{even}}=a_0+a_2+a_4+\cdots+a_{20}.$

Then $P(1)=S_{\text{even}}+S_{\text{odd}},$ $P(-1)=S_{\text{even}}-S_{\text{odd}}.$

Hence, $S_{\text{odd}}=\frac{P(1)-P(-1)}{2}.$

Compute $P(1)$ and $P(-1)$:

$P(1)=(1+1+1)^{10}=3^{10}=59049,$ $P(-1)=(1-1+1)^{10}=1^{10}=1.$

Therefore, $S_{\text{odd}}=\frac{59049-1}{2}=\frac{59048}{2}=29524.$

Find $a_2$, the coefficient of $x^2$ in $(1+x+x^2)^{10}$.

Write $(1+x+x^2)^{10}=\sum \text{terms formed by choosing }1,x,x^2\text{ from each factor.}$

To get total degree $2$, the possibilities are:

• Choose one $x^2$ and the rest $1$: $\binom{10}{1}=10$ ways.

• Choose two $x$'s and the rest $1$: $\binom{10}{2}=45$ ways.

So, $a_2=10+45=55.$

Now compute the required expression:

$S_{\text{odd}}-11a_2=29524-11\cdot 55.$

Since $11\cdot 55=605,$ we get $29524-605=28919.$

Set this equal to $121k$:

$121k=28919.$

Now, $28919\div 121=239,$ because $121\cdot 239=28919.$

Therefore, $k=239.$

Comparison with stored answer:

The derived answer is $239$, which matches the stored correct answer.