If the number of terms in the expansion of (1−2x+4x2)n, x≠0, is 28, then the sum of the coefficients of all the terms in this expansions, is
Clearly, number of terms in the expansion of
(1−2x+4x2)n is (n+2)(n+1)2 or n+2C2
[assuming 1x and 1x2 distinct]
∴ (n+2)(n+1)2=28
⇒ (n + 2)(n + 1) = 56 =( 6 + 1) (6 + 2) ⇒ n = 6
Hence, sum of coefficients =(1−2+4)6=36=729