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JEE Main 20253 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2025 · 3 April · Evening Shift · Mathematics · Binomial Theorem

Let (1+x+x^2)^ 10 =a_0+a_1 x+a_2 x^2+ .+a_ 20 x^ 20 . If...

Let (1+x+x2)10=a0+a1x+a2x2+….+a20x20\left(1+x+x^2\right)^{10}=a_0+a_1 x+a_2 x^2+\ldots .+a_{20} x^{20}. If \left(a_1+a_3+a_5+\ldots .+a_{19}\right)-11 \mathrm{a}_2=121 \mathrm{k},thenkisequalto, then k is equal to\qquad$ .

Correct answer

0

Step-by-step solution

(1+x+x2)10=a0+a1x+a2x2+….+a20x20∴310=a0+a1+a2+….+a20\left(1+x+x^2\right)^{10}=a_0+a_1 x+a_2 x^2+\ldots .+a_{20} x^{20} \therefore 3^{10}=\mathrm{a}_0+\mathrm{a}_1+\mathrm{a}_2+\ldots .+\mathrm{a}_{20} ...(i) 1=a0−a1+a2…..+a201=a_0-a_1+a_2 \ldots . .+a_{20} ...(ii)  (i) - (ii) ⇒a1+a3+….+a19=310−12=29524\text{ (i) - (ii) } \Rightarrow a_1+a_3+\ldots .+a_{19}=\frac{3^{10}-1}{2}=29524 Also {1+x(1+x)}10=1+10C1x(1+x)+10C2x2(1+x)2+….\begin{aligned} \text{ Also }\{ & 1+\mathrm{x}(1+\mathrm{x})\}^{10}=1 \\ & +{ }^{10} \mathrm{C}_1 \mathrm{x}(1+\mathrm{x})+{ }^{10} \mathrm{C}_2 \mathrm{x}^2(1+\mathrm{x})^2+\ldots . \end{aligned}∴a2=10C1+10C2=55∴(a1+a3+…+a19)−11a2121=239\therefore \mathrm{a}_2={ }^{10} \mathrm{C}_1+{ }^{10} \mathrm{C}_2=55 \therefore \frac{\left(\mathrm{a}_1+\mathrm{a}_3+\ldots+\mathrm{a}_{19}\right)-11 \mathrm{a}_2}{121}=239
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