JEE Main 2025 — Binomial Theorem Question with Solution

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

The sum of odd coefficients: $S_{\text{odd }}=a_1+a_3+a_5+\cdots$ $+a_{19}$

Subtracting $11a_2$ from above will give the answer

$$\begin{array}{rl}& S_{\text{odd }}=\frac{f(1)-f(-1)}{2}\\ & f(1)=(1+1+1{)}^{10}=3^{10}\\ & f(-1)=(1-1+1{)}^{10}=(1{)}^{10}=1\\ & S_{\text{odd }}=\sum _{\text{odd }r}{a}_r=\frac{3^{10}-1}{2}\end{array}$$

Now for $a_2$

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

Now use:

$$\begin{array}{rl}& {\left(1-x^3\right)}^{10}=\sum _{x=0}^{10}(-1{)}^k(\genfrac{}{}{0ex}{}{10}{k})x^{3k}\\ & (1-x{)}^{-10}=\sum _{r=0}^{\infty }(\genfrac{}{}{0ex}{}{r+9}{9})x^r\end{array}$$

So

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

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 ${\left(1-x^3\right)}^{10}$ : the $x^0$ term is $\left(\genfrac{}{}{0ex}{}{10}{0}\right)=1$

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

$$\left(\genfrac{}{}{0ex}{}{2+9}{9}\right)=\left(\genfrac{}{}{0ex}{}{11}{9}\right)=55$$

Hence, $a_2=1.55=55$

$$\begin{array}{rl}& \text{ Now, }{S}_{\text{odd }}-11a_2=\frac{3^{10}-1}{2}-11\cdot 55=121k\\ & 3^{10}=59049\end{array}$$

So:

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

So:

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