The number of positive integers k such that the constant term in the binomial expansion of (2 x3+(3/xk))12, x ≠ 0 is 28 ⋅ ℓ, where ℓ is an odd integer, is.

Question

The number of positive integers k such that the constant term in the binomial expansion of (2 x3+(3/xk))12, x ≠ 0 is 28 ⋅ ℓ, where ℓ is an odd integer, is. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

(2 x3+(3/xk))12 t r +1= 12 C r (2 x 3) r ((3/ x k ))12- r x 3 r -(12- r ) k arrow constant ∴ 3 r -12 k + rk =0 ⇒ k =(3 r /12- r ) ∴ possible values of r are 3,6,8,9,10 and corresponding values of k are 1,3,6,9,15 Now 12 C r =220,924,495,220,66 ∴ possible values of k for which we will get 28 are 3,6

Q. The number of positive integers k such that the constant term in the binomial expansion of (2x3+xk3​)12,x=0 is 28⋅ℓ, where ℓ is an odd integer, is_______.

(2x3+xk3​)12 tr+1​=12Cr​(2x3)r(xk3​)12−r x3r−(12−r)k→ constant ∴3r−12k+rk=0 ⇒k=12−r3r​ ∴ possible values of r are 3,6,8,9,10 and corresponding values of k are 1,3,6,9,15 Now 12Cr​=220,924,495,220,66 ∴ possible values of k for which we will get 28 are 3,6

Q. The number of positive integers $k$ such that the constant term in the binomial expansion of $(2 x^{3} + \frac{3}{x ^{k}})^{12}, x \neq = 0$ is $2^{8} \cdot ℓ$ , where $ℓ$ is an odd integer, is_______.