1.

If (3+x2020+x2021)2022=a0+a1x+a2x2+⋯+a10xn, then the value of a0−12a1−12a2+a3−12a4−12a5+a6⋯ is(correct answer + 2, wrong answer - 0.50)

Answer»

If (3+x2020+x2021)2022=a0+a1x+a2x2++a10xn, then the value of a012a112a2+a312a412a5+a6 is

(correct answer + 2, wrong answer - 0.50)



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