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

To solve the problem, we need to determine the value of \( r \) such that \( x^{22} \) is part of the \( (r + 1)^{\text{th}} \) term of the binomial expansion of \((3x^3 - x^2)^9\).

The general formula for the \( (k+1)^{\text{th}} \) term in the binomial expansion of \((a + b)^n\) is:

\(T_{k+1} = \binom{n}{k} a^{n-k} b^k\)

Applying this to our expression \((3x^3 - x^2)^9\), we have:

• \(a = 3x^3\),
• \(b = -x^2\),
• \(n = 9\).

Thus, the general term is:

\(T_{k+1} = \binom{9}{k} (3x^3)^{9-k} (-x^2)^k\)

Simplifying the expression, we have:

\(T_{k+1} = \binom{9}{k} \cdot 3^{9-k} \cdot (x^3)^{9-k} \cdot (-1)^k \cdot (x^2)^k\)

Combining the powers of \( x \), we obtain:

\(T_{k+1} = \binom{9}{k} \cdot 3^{9-k} \cdot (-1)^k \cdot x^{3(9-k) + 2k}\)

Thus, the power of \( x \) in the term \( T_{k+1} \) is:

\(3(9-k) + 2k = 27 - 3k + 2k = 27 - k\)

We are given that the exponent of \( x \) should equal 22, so we set up the equation:

\(27 - k = 22\)

Solving for \( k \):

\(27 - k = 22 \Rightarrow k = 27 - 22 = 5\)

The value of \( r \) is \( k \), which means \( r = 5 \).

Thus, the correct answer is 5.