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

Question

Let ${(1+x+{x}^{2})}^{10}={a}_{0}+{a}_{1}x+{a}_{2}{x}^{2}+\ldots +{a}_{20}{x}^{20}$. If $({a}_{1}+{a}_{3}+{a}_{5}+\ldots +{a}_{19})-11{a}_{2}=121k$, then $k$ is equal to\_\_\_\_\_\_\_\_\_ .

Answer: 239

Accepted Answer

Let $f(x)={(1+x+{x}^{2})}^{10}=\sum _{20}^{r=0}{a}_{r}{x}^{r}$

The sum of odd coefficients: ${S}_{}={a}_{1}+{a}_{3}+{a}_{5}+\cdots$ $+{a}_{19}$

Subtracting $11{a}_{2}$ from above will give the answer

$\begin{gathered} {S}_{}=\frac{f(1)-f(-1)}{2} \ f(1)=(1+1+1{)}^{10}={3}^{10} \ f(-1)=(1-1+1{)}^{10}=(1{)}^{10}=1 \ {S}_{}=\sum _{r}{a}_{r}=\frac{{3}^{10}-1}{2} \end{gathered}$

Now for ${a}_{2}$

$1+x+{x}^{2}=\frac{1-{x}^{3}}{1-x}\Rightarrow f(x)=(\frac{1-{x}^{3}}{1-x})=\frac{{(1-{x}^{3})}^{10}}{(1-x{)}^{10}}$

Now use:

$\begin{gathered} {(1-{x}^{3})}^{10}=\sum _{x=0}^{10}(-1{)}^{k}(\frac{10}{k}){x}^{3k} \ (1-x{)}^{-10}=\sum _{r=0}^{\infty}(\frac{r+9}{9}){x}^{r} \end{gathered}$

So

$f(x)=(\sum _{10}^{k=0}(-1{)}^{k}(\frac{10}{k}){x}^{3k})\cdot (\sum _{\infty}^{r=0}(\frac{r+9}{9}){x}^{r})$

Only the term with ${x}^{0}$ from the first sum (i.e., $k=0$ ) can contribute to ${x}^{2}$, since all other $k\ge 1$ gives ${x}^{3k}\ge$ ${x}^{3}$

From ${(1-{x}^{3})}^{10}$ : the ${x}^{0}$ term is $(\frac{10}{0})=1$

From $(1-x{)}^{-10}$ : the coefficient of ${x}^{2}$ is

$(\frac{2+9}{9})=(\frac{11}{9})=55$

Hence, ${a}_{2}=1.55=55$

$\begin{gathered} {S}_{}-11{a}_{2}=\frac{{3}^{10}-1}{2}-11\cdot 55=121k \ {3}^{10}=59049 \end{gathered}$

So:

$\begin{gathered} S=\frac{59049-1}{2}-605=\frac{59048}{2}-605 \ =29524-605=28919 \end{gathered}$

So:

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

Solutions · 1