Question

If x²² is in the (r+1)th term of the binomial expansion of (3x³-x²)⁹, then the value of r is equal to

• A. 3
• B. 5
• C. 4
• D. 6
• E. 7

Answer 1

The Correct Option is C

We are given the expression \((3x^3 - x^2)^9\) and need to find the value of \(r\) such that \(x^{22}\) appears in the \((r+1)\)th term of its binomial expansion.

The binomial expansion of \((a + b)^n\) is given by: 

\(\sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)

In this case, \(a = 3x^3\) and \(b = -x^2\), and we are expanding \((3x^3 - x^2)^9\). The general term in the expansion is:

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

We need to simplify this expression. The powers of \( x \) in the terms are:

\((3x^3)^{9-k} = 3^{9-k} x^{3(9-k)}\)

\((-x^2)^k = (-1)^k x^{2k}\)

The power of \( x \) in the general term is:

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

We are interested in the term where the power of \( x \) is 22, so we set:

\(27 - k = 22\)

Solving for \( k \), we get:

\(k = 5\)

The term corresponding to \( k = 5 \) is the \((r+1)\)th term. Therefore, \(r = 4\).

The value of \( r \) is 4.