JEE Mains · Maths · STD 11 - 7. binomial theoram
Let \(\left(1+x+x^2\right)^{10}=a_0+a_1 x+a_2 x^2+\ldots .+a_{20} x^{20}\). If \(\left(a_1+a_3+a_5+\ldots .+a_{19}\right)-11 \mathrm{a}_2=121 \mathrm{k}\), then k is equal to ________.
- A 245
- B 239
- C 210
- D 200
Answer & Solution
Correct Answer
(B) 239
Step-by-step Solution
Detailed explanation
\(\left(1+x+x^2\right)^{10}=a_0+a_1 x+a_2 x^2+\ldots .+a_{20} x^{20}\) \(\therefore 3^{10}=\mathrm{a}_0+\mathrm{a}_1+\mathrm{a}_2+\ldots .+\mathrm{a}_{20}\) ...(i) \(1=a_0-a_1+a_2 \ldots . .+a_{20}\) ...(ii)…
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