Let alpha be the constant term in the binomial expansion of

Question

Let alpha be the constant term in the binomial expansion of (sqrt(x)-frac6x^(3 / 2))^n, n <= 15. If the sum of the coefficients of the remaining terms in the expansion is 649 and the coefficient of x^-n is lambda alpha, then lambda is equal to _____________.

DivineJEE · Accepted Answer

Given expression (sqrt(x)-frac6x^3/2)^n. Here, a = sqrt(x) and b = -frac6x^3/2. The r-th term of the binomial expansion of (a+b)^n is given by T_r = ^nC_ra^n-rb^r. Substitute a and b in this formula, we get: T_r = ^nC_r(sqrt(x))^n-r(-frac6x^3/2)^r = ^nC_r(-6)^r x^(n-4r / 2). The constant term in the binomial expansion is obtained when the power of x in the terms equals zero. This happens when (n-4r / 2) = 0, which gives n = 4r. ^n C_(n / 4)(-6)^(n / 4)=alpha \ (-5)^n- ^n C_(n / 4)(-6)^n / 4=649 \ By observation (625 + 24 = 649) , we get n = 4 \ Therefore alpha=-24 Now, for coefficient of x^-4 (n-4 r / 2)=-4 \ n=4 r-8 => r=3 \ lambda(-24)=(-6)^3 * ^4 C_3 \ => lambda=36

Explanation

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