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 $$ x $$

UpStudy AI · Accepted Answer

**i. Finding the $$6^{\text{th}}$$ term**

The general term in the expansion of $$(3x-2)^{10}$$ is given by  
$$
T_{k+1} = \binom{10}{k}(3x)^{10-k}(-2)^k,
$$  
with $$ k = 0, 1, 2, \ldots, 10 $$.

The $$6^{\text{th}}$$ term corresponds to $$ k=5 $$. Thus,
$$
T_6 = \binom{10}{5}(3x)^{10-5}(-2)^5.
$$

Calculate each part:
- $$\binom{10}{5} = 252$$,
- $$(3x)^5 = 3^5x^5 = 243x^5$$,
- $$(-2)^5 = -32$$.

Now, multiply them:
$$
T_6 = 252 \cdot 243x^5 \cdot (-32) = -1,\!959,\!552\,x^5.
$$

---

**ii. Finding the coefficient of the $$4^{\text{th}}$$ term**

The $$4^{\text{th}}$$ term in the expansion corresponds to $$ k=3 $$ (since the term number is $$ k+1 $$). Hence,
$$
T_4 = \binom{10}{3}(3x)^{10-3}(-2)^3.
$$

Compute each factor:
- $$\binom{10}{3} = 120$$,
- $$(3x)^7 = 3^7x^7 = 2187x^7$$,
- $$(-2)^3 = -8$$.

Thus,
$$
T_4 = 120 \cdot 2187x^7 \cdot (-8).
$$

The coefficient is:
$$
120 \times 2187 \times (-8) = -2,\!099,\!520.
$$

---

**iii. Finding the third term in ascending powers of $$x$$**

In the expansion $$(3x-2)^{10}$$, the terms are naturally written in descending powers of $$x$$. For ascending powers of $$x$$, they are rearranged so that the constant term (with $$ x^0 $$) comes first.

Notice that in the general term,
$$
T_{k+1} = \binom{10}{k}(3x)^{10-k}(-2)^k,
$$
the power of $$ x $$ is $$10-k$$. For ascending order (starting with the lowest power), we want:
- 1st term: $$ x^0 $$ $$ \Rightarrow 10-k = 0 $$ which gives $$ k=10 $$,
- 2nd term: $$ x^1 $$ $$ \Rightarrow 10-k = 1 $$ which gives $$ k=9 $$,
- 3rd term: $$ x^2 $$ $$ \Rightarrow 10-k = 2 $$ which gives $$ k=8 $$.

Thus, the third term (in ascending powers) corresponds to $$ k=8 $$:
$$
T = \binom{10}{8}(3x)^{10-8}(-2)^8.
$$

Now, compute:
- $$\binom{10}{8} = \binom{10}{2} = 45$$,
- $$(3x)^2 = 9x^2$$,
- $$(-2)^8 = 256$$.

Multiply the factors:
$$
T = 45 \cdot 9x^2 \cdot 256 = 103680\,x^2.
$$

---

**Final Answers:**

i. $$ -1,\!959,\!552\,x^5 $$  
ii. The coefficient of the $$4^{\text{th}}$$ term is $$-2,\!099,\!520$$.  
iii. The third term in ascending powers of $$x$$ is $$ 103680\,x^2 $$.
i. The 6th term is $$-1,\!959,\!552\,x^5$$. ii. The coefficient of the 4th term is $$-2,\!099,\!520$$. iii. The third term in ascending powers of $$x$$ is $$103,\!680\,x^2$$.

B. In the expansion of , what is the
i. term
ii. coefficient of the term
iii. third term in ascending powers of

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