Question

If $x^{22}$ is in the $(r + 1)^{\text{th}}$ term of the binomial expansion of $(3x^3 - x^2)^9$, then the value of $r$ is equal to

• A. 3
• B. 4
• C. 5
• D. 6
• E. 7

Hint:

Always check the sign. $(-x^2)^r$ will affect the sign of the term but not the power of $x$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We must find the specific term in a binomial expansion that contains a given power of \(x\).\ The general formula for the \( (r+1)^{\text{th}} \) term in the expansion of \( (a + b)^n \) is used.\ Step 2: Key Formula or Approach:
The general term \(T_{r+1}\) is given by: \[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \] Here, \( a = 3x^3 \), \( b = -x^2 \), and \( n = 9 \). Step 3: Detailed Explanation:
Substitute the given terms into the general formula: \[ T_{r+1} = \binom{9}{r} (3x^3)^{9-r} (-x^2)^r \] Now, separate the constants from the variables: \[ T_{r+1} = \binom{9}{r} \left(3^{9-r}\right) \left(x^{3(9-r)}\right) (-1)^r \left(x^{2r}\right) \] \[ T_{r+1} = \binom{9}{r} 3^{9-r} (-1)^r x^{27 - 3r} x^{2r} \] Combine the exponents of \(x\) by adding them together: \[ \text{Power of } x = (27 - 3r) + 2r = 27 - r \] The problem states that this term must contain \(x^{22}\).\ Therefore, set the combined exponent of \(x\) equal to 22: \[ 27 - r = 22 \] Solve for \(r\): \[ r = 27 - 22 = 5 \] Step 4: Final Answer:
The value of \(r\) is 5.