© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the General Term of the Binomial Expansion
We have the expression
\left(2x^3 + \frac{3}{x^k}\right)^{12}.
The general term (the (r+1) th term) in its binomial expansion is given by:
T_{r+1} = \binom{12}{r} \left(2x^3\right)^r \left(\frac{3}{x^k}\right)^{12 - r}.
Step 2: Simplify the General Term
Simplifying inside the power of x , we get:
T_{r+1}
= \binom{12}{r} \, 2^r \, 3^{12-r} \, x^{3r} \, x^{-k(12-r)}
= \binom{12}{r} \, 2^r \, 3^{12-r} \, x^{3r - k(12-r)}.
Hence, the net exponent of x in the general term is
3r - k(12-r).
Step 3: Impose the Constant Term Condition
For the term to be independent of x , the exponent of x must be zero:
3r - k(12-r) = 0 \quad \Longrightarrow \quad 3r = k(12-r) \quad \Longrightarrow \quad k = \frac{3r}{12-r}.
We seek integer values of k>0 for various integer r (where 0 \le r \le 12 ).
Step 4: Find Possible Integer Values of k
By substituting r = 0, 1, 2, \dots, 12 in k = \frac{3r}{12-r} and checking for positive integers, we get:
r = 3 \,\Rightarrow\, k = 1.
r = 6 \,\Rightarrow\, k = 3.
r = 8 \,\Rightarrow\, k = 6.
r = 9 \,\Rightarrow\, k = 9.
r = 10 \,\Rightarrow\, k = 15.
r = 11 \,\Rightarrow\, k = 33.
So the integer values of k are 1, 3, 6, 9, 15, and 33.
Step 5: Check the Form of the Constant Term for 2^8 \cdot l (with l Odd)
We need the constant term to be exactly of the form 2^8 \cdot l, where l is an odd integer. We substitute each valid k above back into the constant term expression and check the power of 2 in that term:
For k = 1 (r = 3):
The constant term becomes
\binom{12}{3} 2^3 3^9.
Factoring out powers of 2 shows it leads to 2^5 \times \text{(some integer)}, so the power of 2 is 5, which is not 2^8. Fails the condition.
For k = 3 (r = 6):
The constant term is
\binom{12}{6} 2^6 3^6.
Detailed calculation shows it factors out exactly 2^8 \cdot (\text{odd integer}), so this satisfies the requirement.
For k = 6 (r = 8):
The constant term is
\binom{12}{8} 2^8 3^4.
On simplification, it again has exactly 2^8 as a factor, and what remains is an odd integer. Satisfies the requirement.
For k = 9 (r = 9):
The constant term
\binom{12}{9} 2^9 3^3
gives a factor of 2^{11} overall when fully expanded, making the remaining factor even. This does not match 2^8 \cdot l with l odd. Fails the condition.
For k = 15 (r = 10):
Similar analysis shows the power of 2 exceeds 8, making l even. Fails the condition.
For k = 33 (r = 11):
Again, the exact power of 2 is not 8 (it surpasses 8), so l becomes even. Fails the condition.
Step 6: Conclude the Number of Valid k
Only k = 3 and k = 6 yield a constant term of the form 2^8 \cdot l with l odd. Hence, there are 2 such positive integers k.
