The sum of the first $$ n $$ term of a series is given by $$ S_{n}=2^{n+2}-4 $$ Determine the sum of tle fiost 24 terms Determine the $$ 24^{ ext {th }} $$ term. 3 Prove thiat the $$ n^{ ext {th }} $$ term of the seres is $$ 2^{n+1} $$

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The sum of the first $$ n $$ term of a series is given by $$ S_{n}=2^{n+2}-4 $$ Determine the sum of tle fiost 24 terms Determine the $$ 24^{	ext {th }} $$ term. 3 Prove thiat the $$ n^{	ext {th }} $$ term of the seres is $$ 2^{n+1} $$

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Certainly! Let's address each part of your question systematically.

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### **Given:**
The sum of the first $$ n $$ terms of a series is:
$$
S_{n} = 2^{n+2} - 4
$$

---

### **1. Determine the Sum of the First 24 Terms ($$ S_{24} $$)**

To find the sum of the first 24 terms, substitute $$ n = 24 $$ into the formula for $$ S_n $$:

$$
S_{24} = 2^{24+2} - 4 = 2^{26} - 4
$$

**Calculating $$ 2^{26} $$:**
$$
2^{10} = 1,\!024
$$
$$
2^{20} = 1,\!048,\!576 \quad (\text{since } 2^{20} = (2^{10})^2)
$$
$$
2^{26} = 2^{20} \times 2^{6} = 1,\!048,\!576 \times 64 = 67,\!108,\!864
$$

Now, subtract 4:
$$
S_{24} = 67,\!108,\!864 - 4 = 67,\!108,\!860
$$

**$$\boxed{S_{24} = 67,\!108,\!860}$$**

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### **2. Determine the 24th Term ($$ a_{24} $$)**

The $$ n^{\text{th}} $$ term of a series can be found using the difference between consecutive sums:
$$
a_{n} = S_n - S_{n-1}
$$

So, for the 24th term:
$$
a_{24} = S_{24} - S_{23}
$$

First, find $$ S_{23} $$:
$$
S_{23} = 2^{23+2} - 4 = 2^{25} - 4 = 33,\!554,\!432 - 4 = 33,\!554,\!428
$$

Now, compute $$ a_{24} $$:
$$
a_{24} = 67,\!108,\!860 - 33,\!554,\!428 = 33,\!554,\!432
$$

Alternatively, using the formula derived in part 3 ($$ a_n = 2^{n+1} $$):
$$
a_{24} = 2^{24+1} = 2^{25} = 33,\!554,\!432
$$

**$$\boxed{a_{24} = 33,\!554,\!432}$$**

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### **3. Prove that the $$ n^{\text{th}} $$ Term of the Series is $$ 2^{n+1} $$**

To find a general expression for the $$ n^{\text{th}} $$ term ($$ a_n $$), consider the sum of the first $$ n $$ terms and the first $$ n-1 $$ terms:

$$
a_n = S_n - S_{n-1}
$$

Given:
$$
S_n = 2^{n+2} - 4
$$
$$
S_{n-1} = 2^{(n-1)+2} - 4 = 2^{n+1} - 4
$$

Subtract $$ S_{n-1} $$ from $$ S_n $$:
$$
a_n = (2^{n+2} - 4) - (2^{n+1} - 4) = 2^{n+2} - 4 - 2^{n+1} + 4 = 2^{n+2} - 2^{n+1}
$$

Factor out $$ 2^{n+1} $$:
$$
a_n = 2^{n+1} (2 - 1) = 2^{n+1} \times 1 = 2^{n+1}
$$

**Therefore, the $$ n^{\text{th}} $$ term is $$ \boxed{2^{n+1}} $$.**

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### **Summary**

1. **Sum of the first 24 terms:** $$ S_{24} = 67,\!108,\!860 $$
2. **24th term:** $$ a_{24} = 33,\!554,\!432 $$
3. **General term:** $$ a_n = 2^{n+1} $$

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If you have any further questions or need additional clarification on any of these steps, feel free to ask!
The sum of the first 24 terms is 67,108,860. The 24th term is 33,554,432. The nth term of the series is $$2^{n+1}$$.

The sum of the first \( n \) term of a series is given by \( S_{n}=2^{n+2}-4 \) Determine the sum of tle fiost 24 terms Determine the \( 24^{\text {th }} \) term. 3 Prove thiat the \( n^{\text {th }} \) term of the seres is \( 2^{n+1} \)

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