JEE Mathematics 2025 Question with Solution

Given:

$$(1 + x + x^2)^{10} = a_0 + a_1 x + a_2 x^2 + \dots + a_{20} x^{20}$$

Find: $k$, where

$$(a_1 + a_3 + a_5 + \dots + a_{19}) - 11a_2 = 121k$$

Let

$$P(x) = (1 + x + x^2)^{10}$$

To find the sum of coefficients of odd powers, use

$$\frac{P(1) - P(-1)}{2}$$

Now,

$$P(1) = (1 + 1 + 1)^{10} = 3^{10}$$

and

$$P(-1) = (1 - 1 + (-1)^2)^{10} = 1^{10} = 1$$

Therefore,

$$a_1 + a_3 + a_5 + \dots + a_{19} = \frac{3^{10} - 1}{2} = \frac{59049 - 1}{2} = \frac{59048}{2} = 29524$$

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

$$(1 + x + x^2)^{10} = (1 + (x + x^2))^{10}$$

Using binomial expansion, only the terms with powers up to $2$ in $x$ contribute to $a_2$:

$$\binom{10}{1}(x + x^2)$$

and

$$\binom{10}{2}(x + x^2)^2$$

Now,

$$\binom{10}{1}(x + x^2) = 10x + 10x^2$$

so the coefficient of $x^2$ from this term is $10$. Also,

$$\binom{10}{2}(x + x^2)^2 = 45(x^2 + 2x^3 + x^4)$$

so the coefficient of $x^2$ from this term is $45$. Hence,

$$a_2 = 10 + 45 = 55$$

Substitute into the given relation:

$$29524 - 11(55) = 121k$$ $$29524 - 605 = 121k$$ $$28919 = 121k$$ $$k = \frac{28919}{121} = 239$$

Therefore, the value of $k = 239$.

Given:

$$P(x) = (1 + x + x^2)^{10} = \sum_{r=0}^{20} a_r x^r$$

Find: $k$ from

$$(a_1 + a_3 + a_5 + \dots + a_{19}) - 11a_2 = 121k$$

First compute the sum of odd-power coefficients. From

$$P(1) = a_0 + a_1 + a_2 + a_3 + \dots + a_{20} = 3^{10}$$

and

$$P(-1) = a_0 - a_1 + a_2 - a_3 + \dots + a_{20} = 1$$

Subtracting,

$$P(1) - P(-1) = 2(a_1 + a_3 + a_5 + \dots + a_{19})$$

So,

$$2(a_1 + a_3 + a_5 + \dots + a_{19}) = 3^{10} - 1 = 59049 - 1 = 59048$$

Thus,

$$a_1 + a_3 + a_5 + \dots + a_{19} = 29524$$

Now compute $a_2$. In

$$(1 + (x + x^2))^{10} = \binom{10}{0} + \binom{10}{1}(x + x^2) + \binom{10}{2}(x + x^2)^2 + \dots$$

the coefficient of $x^2$ comes only from:

1. $\binom{10}{1}(x + x^2)$, contributing $10$
2. $\binom{10}{2}(x + x^2)^2$, where

$$(x + x^2)^2 = x^2 + 2x^3 + x^4$$

so the contribution is $\binom{10}{2} = 45$

Therefore,

$$a_2 = 10 + 45 = 55$$

Now use the given equation:

$$(a_1 + a_3 + a_5 + \dots + a_{19}) - 11a_2 = 121k$$ $$29524 - 11 \times 55 = 121k$$ $$29524 - 605 = 121k$$ $$28919 = 121k$$ $$k = 239$$

Therefore, the value of $k$ is $239$.