The number of positive integers k such that the constant term in the binomial expansion of \(( 2x^3 + \frac {3}{x^k} )^{12}, x ≠ 0\) is \(2^8\) . l, where l is an odd integer, is______.
Correct Answer: 2
Solution and Explanation
\(T_{r+1} = ^{12}C_r (2x^3)^{12-r}( \frac {3}{x^k })^r\)
\(T_{r+1}= ^{12}C_r2^{12-r}3^r X^{36-3r-kr}\)
For constant term \(36 – 3r – kr = 0\)
\(r = \frac {36}{3+k}\)
So, k can be \(1, 3, 6, 9, 15, 33\)
In order to get \(2^8\), check by putting values of k and corresponding in general term. By checking, it is possible only where k = \(3\) or \(6\)
So, the answer is \(2\).
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Concepts Used:
Binomial Theorem
The binomial theorem formula is used in the expansion of any power of a binomial in the form of a series. The binomial theorem formula is
Properties of Binomial Theorem
- The number of coefficients in the binomial expansion of (x + y)n is equal to (n + 1).
- There are (n+1) terms in the expansion of (x+y)n.
- The first and the last terms are xn and yn respectively.
- From the beginning of the expansion, the powers of x, decrease from n up to 0, and the powers of a, increase from 0 up to n.
- The binomial coefficients in the expansion are arranged in an array, which is called Pascal's triangle. This pattern developed is summed up by the binomial theorem formula.