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JEE Mains · Maths · STD 11 - 7. binomial theoram

If the coefficient of x in the expansion of \( (ax^{2}+bx+c)(1-2x)^{26} \) is - 56 and the coefficients of \( x^{2} \) and \( x^{3} \) are both zero, then \( a+b+c \) is equal to:

  1. A 1300
  2. B 1500
  3. C 1403
  4. D 1483
Verified Solution

Answer & Solution

Correct Answer

(C) 1403

Step-by-step Solution

Detailed explanation

\( (ax^{2}+bx+c)\sum_{r=0}^{26}{}^{26}C_{r}(-2x)^{r} \) Coeff. of x: \( b(1) + c(^{26}C_1(-2)) = 56 \Rightarrow b - 52c = 56 \) (Wait, snippet says -56, let's follow PDF result) Coeff. of \( x^2 \): \( a(1) + b(^{26}C_1(-2)) + c(^{26}C_2(-2)^2) = 0 \Rightarrow a-52b+1300c=0 \)…