1) $$ \frac{ ext { Simplify }}{2 x+3} $$ $$ \frac{2^{2 x+3}-5.2^{2 x+1}}{4^{x+2}} $$ $$ \frac{1,2\left(a b^{2} c ight)^{3}}{a^{-1} b^{4} c^{3}} imes \frac{\left(a^{-2} b^{3} c^{5} ight)^{0}}{a^{-1} b^{4} c^{2}} $$ 2) Arithmetic Sequence $$ \& $$ deries 2.1 Given $$ S_{n}=2970 $$, and the Sequence $$ 12 ; 18 ; 24 \ldots $$. Find number of terms ( $$ n $$ )

Question

1) $$ \frac{	ext { Simplify }}{2 x+3} $$ $$ \frac{2^{2 x+3}-5.2^{2 x+1}}{4^{x+2}} $$ $$ \frac{1,2\left(a b^{2} cight)^{3}}{a^{-1} b^{4} c^{3}} 	imes \frac{\left(a^{-2} b^{3} c^{5}ight)^{0}}{a^{-1} b^{4} c^{2}} $$ 2) Arithmetic Sequence $$ \& $$ deries 2.1 Given $$ S_{n}=2970 $$, and the Sequence $$ 12 ; 18 ; 24 \ldots $$. Find number of terms ( $$ n $$ )

UpStudy AI · Accepted Answer

Simplify the expression by following steps:
- step0: Solution:
  $$\frac{2^{2x+3}-5\times 2^{2x+1}}{4^{x+2}}$$
- step1: Subtract the terms:
  $$\frac{-2^{2x+1}}{4^{x+2}}$$
- step2: Calculate:
  $$\frac{-2^{2x+1}}{\left(2^{2}\right)^{x+2}}$$
- step3: Calculate:
  $$\frac{-2^{2x+1}}{2^{2x+4}}$$
- step4: Calculate:
  $$-2^{-3}$$
Solve the equation $$ S_n=\frac{n}{2}(2a+(n-1)d) $$.
Solve the equation by following steps:
- step0: Solve for $$n$$:
  $$S_{n}=\frac{n}{2}\left(2a+\left(n-1\right)d\right)$$
- step1: Simplify:
  $$S_{n}=\frac{n\left(2a+dn-d\right)}{2}$$
- step2: Rewrite the expression:
  $$S_{n}=\frac{n\left(2a-d+dn\right)}{2}$$
- step3: Swap the sides:
  $$\frac{n\left(2a-d+dn\right)}{2}=S_{n}$$
- step4: Cross multiply:
  $$n\left(2a-d+dn\right)=2S_{n}$$
- step5: Expand the expression:
  $$2an-dn+dn^{2}=2S_{n}$$
- step6: Collect like terms:
  $$\left(2a-d\right)n+dn^{2}=2S_{n}$$
- step7: Move the expression to the left side:
  $$\left(2a-d\right)n+dn^{2}-2S_{n}=0$$
- step8: Rewrite in standard form:
  $$dn^{2}+\left(2a-d\right)n-2S_{n}=0$$
- step9: Solve using the quadratic formula:
  $$n=\frac{-2a+d\pm \sqrt{\left(2a-d\right)^{2}-4d\left(-2S_{n}\right)}}{2d}$$
- step10: Simplify the expression:
  $$n=\frac{-2a+d\pm \sqrt{4a^{2}-4ad+d^{2}+8dS_{n}}}{2d}$$
- step11: Separate into possible cases:
  $$\begin{align}&n=\frac{-2a+d+\sqrt{4a^{2}-4ad+d^{2}+8dS_{n}}}{2d}\&n=\frac{-2a+d-\sqrt{4a^{2}-4ad+d^{2}+8dS_{n}}}{2d}\end{align}$$
Let's solve the problems step by step.

### 1) Simplification

#### Expression 1:
We need to simplify the expression:
$$
\frac{2^{2x+3} - 5 \cdot 2^{2x+1}}{4^{x+2}}
$$

First, we can rewrite $$4^{x+2}$$ as $$(2^2)^{x+2} = 2^{2(x+2)} = 2^{2x+4}$$.

Now, the expression becomes:
$$
\frac{2^{2x+3} - 5 \cdot 2^{2x+1}}{2^{2x+4}}
$$

Next, we can factor out $$2^{2x+1}$$ from the numerator:
$$
= \frac{2^{2x+1}(2^2 - 5)}{2^{2x+4}} = \frac{2^{2x+1}(-1)}{2^{2x+4}} = -\frac{2^{2x+1}}{2^{2x+4}} = -2^{-3}
$$

So, the simplified result is:
$$
-2^{-3}
$$

#### Expression 2:
Next, we simplify:
$$
\frac{1.2(a b^{2} c)^{3}}{a^{-1} b^{4} c^{3}} \times \frac{(a^{-2} b^{3} c^{5})^{0}}{a^{-1} b^{4} c^{2}}
$$

Since $$(a^{-2} b^{3} c^{5})^{0} = 1$$, the expression simplifies to:
$$
\frac{1.2(a b^{2} c)^{3}}{a^{-1} b^{4} c^{3}} \times \frac{1}{a^{-1} b^{4} c^{2}}
$$

Now, we can combine the fractions:
$$
= 1.2 \cdot \frac{(a b^{2} c)^{3}}{(a^{-1} b^{4} c^{3})^2}
$$

Calculating the numerator:
$$
(a b^{2} c)^{3} = a^{3} b^{6} c^{3}
$$

Calculating the denominator:
$$
(a^{-1} b^{4} c^{3})^2 = a^{-2} b^{8} c^{6}
$$

Now, the expression becomes:
$$
1.2 \cdot \frac{a^{3} b^{6} c^{3}}{a^{-2} b^{8} c^{6}} = 1.2 \cdot a^{3 - (-2)} b^{6 - 8} c^{3 - 6} = 1.2 \cdot a^{5} b^{-2} c^{-3}
$$

### 2) Arithmetic Sequence

Given $$ S_n = 2970 $$ for the sequence $$ 12, 18, 24, \ldots $$, we can identify the first term $$ a = 12 $$ and the common difference $$ d = 6 $$.

The formula for the sum of the first $$ n $$ terms of an arithmetic sequence is:
$$
S_n = \frac{n}{2}(2a + (n-1)d)
$$

Substituting the known values:
$$
2970 = \frac{n}{2}(2 \cdot 12 + (n-1) \cdot 6)
$$

This simplifies to:
$$
2970 = \frac{n}{2}(24 + 6n - 6) = \frac{n}{2}(6n + 18)
$$

Multiplying both sides by 2:
$$
5940 = n(6n + 18)
$$

Rearranging gives:
$$
6n^2 + 18n - 5940 = 0
$$

Dividing the entire equation by 6:
$$
n^2 + 3n - 990 = 0
$$

Now we can solve for $$ n $$ using the quadratic formula:
$$
n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot (-990)}}{2 \cdot 1}
$$

Calculating the discriminant:
$$
= \frac{-3 \pm \sqrt{9 + 3960}}{2} = \frac{-3 \pm \sqrt{3969}}{2} = \frac{-3 \pm 63}{2}
$$

Calculating the two possible values for $$ n $$:
1. $$ n = \frac{60}{2} = 30 $$
2. $$ n = \frac{-66}{2} = -33 $$ (not valid)

Thus, the number of terms $$ n $$ is:
$$
n = 30
$$

### Summary of Results
1. The simplified expression is $$ -2^{-3} $$.
2. The number of terms $$ n $$ in the arithmetic sequence is $$ 30 $$.
1. The simplified expression is $$ -2^{-3} $$. 2. The number of terms $$ n $$ in the arithmetic sequence is $$ 30 $$.

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