AP EAMCET · Maths · Binomial Theorem
Coefficient of \(x^2\) in the expansion of \(\left(x^2+x-2\right)^5\) is
- A 800
- B 756
- C 0
- D 512
Answer & Solution
Correct Answer
(C) 0
Step-by-step Solution
Detailed explanation
\(\frac{5!}{0!2!3!}(1)^0(1)^2(-2)^3 + \frac{5!}{1!0!4!}(1)^1(1)^0(-2)^4\) \(10(1)(1)(-8) + 5(1)(1)(16)\) \(-80 + 80\) \(0\)
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