B. In the expansion of \( (3 x-2)^{10} \), what is the i. \( 6^{\text {th }} \) term ii. coefficient of the \( 4^{\text {th }} \) term iii. third term in ascending powers of \( x \)

**i. Finding the 6th term** The general term in the expansion of \( (3x-2)^{10} \) is given by \[ T_{k+1} = \binom{10}{k}(3x)^{10-k}(-2)^k. \] For the 6th term, we have \( k = 5 \) (since the term number is \( k+1 \)): \[ T_6 = \binom{10}{5}(3x)^{10-5}(-2)^5. \] Calculate each part: - \(\binom{10}{5} = 252.\) - \((3x)^5 = 3^5 x^5 = 243x^5.\) - \((-2)^5 = -32.\) Now, multiply them together: \[ T_6 = 252 \times 243x^5 \times (-32) = -252 \times 243 \times 32 \, x^5. \] First, compute \(252 \times 243 = 61236\). Then, \[ 61236 \times 32 = 1,959,552. \] Thus, the 6th term is \[ T_6 = -1,959,552\, x^5. \] --- **ii. Finding the coefficient of the 4th term** The 4th term corresponds to \( k = 3 \) (since \( k+1=4 \)): \[ T_4 = \binom{10}{3}(3x)^{10-3}(-2)^3. \] Ignore \( x \) to find the coefficient: - \(\binom{10}{3} = 120.\) - \((3x)^7\) contributes a factor of \( 3^7 = 2187.\) - \((-2)^3 = -8.\) Thus, the coefficient of the 4th term is \[ 120 \times 2187 \times (-8). \] First, compute \(120 \times 2187 = 262440\), then multiplying by \(-8\) gives: \[ 262440 \times (-8) = -2,099,520. \] So, the coefficient of the 4th term is \(-2,099,520\). --- **iii. Finding the third term in ascending powers of \( x \)** In the binomial expansion \[ (3x-2)^{10} = \sum_{k=0}^{10}\binom{10}{k}(3x)^{10-k}(-2)^k, \] the exponent of \( x \) in each term is \( 10-k \). Arranging in ascending order of \( x \) (from lowest to highest power), the term with the lowest power \( x^0 \) occurs when \( k=10 \), the next one with \( x^1 \) occurs when \( k=9 \), and so the third term in ascending order corresponds to \( k=8 \). For \( k = 8 \): \[ T = \binom{10}{8}(3x)^{10-8}(-2)^8. \] Calculate each part: - \(\binom{10}{8} = \binom{10}{2} = 45.\) - \((3x)^2 = 9x^2.\) - \((-2)^8 = 256.\) Multiply the factors: \[ 45 \times 9x^2 \times 256. \] First, \(45 \times 9 = 405\). Then, \[ 405 \times 256 = 103680. \] Thus, the third term (in ascending powers of \( x \)) is \[ 103680\, x^2. \]