JEE Main 2025 Apr 3 Shift 2, Mathematics Q22: Introduction to Binomial Expansion

JEE Main2025Apr 3, Shift 2Mathematics

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

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Solution

Set $x = 1$ : $3^{10} = a_{0} + a_{1} + a_{2} + \dots + a_{20}$ .

Set $x = - 1$ : $1 = a_{0} - a_{1} + a_{2} - \dots + a_{20}$ .

Subtract to isolate odd-indexed coefficients:

$a_{1} + a_{3} + \dots + a_{19} = \frac{3 ^{10} - 1}{2} = \frac{59049 - 1}{2} = 29524$ .

Find $a_{2}$ by rewriting $(1 + x + x^{2})^{10} = (1 + x (1 + x))^{10}$ and using the binomial theorem:

$(1 + x (1 + x))^{10} = j = 0 \sum 10 (j 10) x^{j} (1 + x)^{j}$ .

The $x^{2}$ coefficient gets contributions from $j = 1$ (with $x (1 + x)$ providing $x \cdot x$ , coefficient $(1 10) \cdot 1$ ) and $j = 2$ (with $x^{2} (1 + x)^{2}$ providing $x^{2} \cdot 1$ , coefficient $(2 10)$ ):

$a_{2} = (1 10) + (2 10) = 10 + 45 = 55$ .

Therefore $\frac{( a _{1} + a _{3} + \dots + a _{19} ) - 11 a _{2}}{121} = \frac{29524 - 11 \cdot 55}{121} = \frac{29524 - 605}{121} = \frac{28919}{121} = 239$ .

$∴ k = 239$ .

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JEE Main 2025 Apr 3 Shift 2, Mathematics Q22: Introduction to Binomial Expansion

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