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

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

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**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
$$
103680\, x^2.
$$
**i. 6th Term:** $$-1,959,552\, x^5$$ **ii. Coefficient of the 4th Term:** $$-2,099,520$$ **iii. Third Term in Ascending Powers of $$ x $$:** $$103,680\, x^2$$

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 \( x \)

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