JEE Mains · Maths · STD 11 - 7. binomial theoram
If the coefficents of \({x^3}\) and \({x^4}\) in the expansion of \(\left( {1 + ax + b{x^2}} \right){\left( {1 - 2x} \right)^{18}}\) in powers of \(x\) are both zero, then \( (a,b) \) is equal to
- A (\(14\),\(\frac{{272}}{3}\))
- B (\(16\),\(\frac{{272}}{3}\))
- C (\(16\),\(\frac{{251}}{3}\))
- D (\(14\),\(\frac{{251}}{3}\))
Answer & Solution
Correct Answer
(B) (\(16\),\(\frac{{272}}{3}\))
Step-by-step Solution
Detailed explanation
In the expansion of \(\left(1+a x+b x^{2}\right)(1-2 x)^{18},\) Coefficient of \(x^{3}\) in \(\left(1+a x+b x^{2}\right)(1-2 x)^{18}\) \(=\) Coefficient of \(x^{3}\) in \((1-2 x)^{18}\) \({+\text { Coefficient of } x^{2} \text { in a }(1-2 x)^{\text {18 }}}\)…
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