JEE Main 2025 — 3 April, Evening Shift — Question 45

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 11a2 from above will give the answer

$S_{\text {odd }}=\frac{f(1)-f(-1)}{2}$ $f(1)=(1+1+1)^{10}=3^{10}$

$f(-1)=(1-1+1)^{10}=(1)^{10}=1$

$S_{\text {odd }}=\sum_{\text {odd } r} a_{r}=\frac{3^{10}-1}{2}$

Now for $\mathrm{a}_{2}$

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

Now use: $\left(1-x^{3}\right)^{10}=\sum_{x=0}^{10}(-1)^{k}\binom{10}{k} x^{3 k}$

$(1-x)^{-10}=\sum_{r=0}^{\infty}\binom{r+9}{9} x^{r}$

So $f(x)=\left(\sum_{k=0}^{10}(-1)^{k}\binom{10}{k} x^{3 k}\right) \cdot\left(\sum_{r=0}^{\infty}\binom{r+9}{9} x^{r}\right)$

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

$x^{3 k} \geq$ $x^{3}$

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

From $(1-x)^{-10}$ : the coefficient of $x^{2}$ is $\binom{2+9}{9}=\binom{11}{9}=55$

Hence, $a_{2}=1 \cdot 55=55$

Now, $S_{odd}$ $-11 a_{2}=\frac{3^{10}-1}{2}-11.55=121 k$

$3^{10}=59049$

So: $S=\frac{59049-1}{2}-605=\frac{59048}{2}-605$

$=29524-605=28919$

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