(1+x+x2)10=a0+a1x+a2x2+….+a20x20 ∴310=a0+a1+a2+….+a20 ...(i) 1=a0−a1+a2…..+a20 ...(ii) (i) - (ii) ⇒a1+a3+….+a19=2310−1=29524 $\begin{aligned}
\text { Also }{ & 1+\mathrm{x}(1+\mathrm{x})}^{10}=1 \
& +{ }^{10} \mathrm{C}_1 \mathrm{x}(1+\mathrm{x})+{ }^{10} \mathrm{C}_2 \mathrm{x}^2(1+\mathrm{x})^2+\ldots .
\end{aligned}\therefore \mathrm{a}_2={ }^{10} \mathrm{C}_1+{ }^{10} \mathrm{C}_2=55\therefore \frac{\left(\mathrm{a}_1+\mathrm{a}3+\ldots+\mathrm{a}{19}\right)-11 \mathrm{a}_2}{121}=239$
Let (1+x+x^2 )^10=a_0+a_1 x+a_2 x^2+… .+a_20 x^20. If (a_1+a_3+a_5+… .+a_19 )-11 a_2=121 k, then k is equal to .
JEE Main 2025 — Mathematics Algebra
Let (1+x+x2)10=a0+a1x+a2x2+….+a20x20. If (a1+a3+a5+….+a19)−11a2=121k, then k is equal to .
Held on 3 Apr 2025 · Verified 6 Jul 2026.
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