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← JEE Main 2016 · 3 Apr · Shift 1
MathematicsSingle correct+4 / -1Question 67 of 90

JEE Main 2016 · 3 Apr · Shift 1, Mathematics question 67

If the number of terms in the expansion of (1−2x+4x2)n\left(1-\dfrac2x+\dfrac4{x^2}\right)^n, x≠0x\ne0, is 28, then the sum of the coefficients of all the terms in this expansion, is :

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Answer and worked solution

Official answer: D

Treated as a trinomial, the expansion has (n+22)=28\binom{n+2}2=28 terms, so n=6n=6. The sum of the coefficients is the value at x=1x=1: (1−2+4)6=729(1-2+4)^6=729.