3. Calculate: $$ \begin{array}{ll} ext { a) } \quad \sum_{n=0}^{4} 3 n^{2} & ext { b) } \sum_{n=1}^{10} 90-3 n \ ext { c) } \quad \sum_{k=5}^{25} 2 n-1 & ext { d) } \sum_{n=1}^{13} 5\left(\frac{3}{2} ight)^{n-1} \ ext { e) } \quad \sum_{n=1}^{\infty} 3\left(\frac{1}{2} ight)^{n-1} & \ ext { 4. Derive the formula for: } \ ext { a) } S_{n}=\frac{n}{2}[2 a+(n-1) d] \ ext { b) } S_{n}=\frac{a\left(1-r^{n} ight)}{1-r}\end{array} $$

Question

3. Calculate: $$ \begin{array}{ll}	ext { a) } \quad \sum_{n=0}^{4} 3 n^{2} & 	ext { b) } \sum_{n=1}^{10} 90-3 n \ 	ext { c) } \quad \sum_{k=5}^{25} 2 n-1 & 	ext { d) } \sum_{n=1}^{13} 5\left(\frac{3}{2}ight)^{n-1} \ 	ext { e) } \quad \sum_{n=1}^{\infty} 3\left(\frac{1}{2}ight)^{n-1} & \ 	ext { 4. Derive the formula for: } \ 	ext { a) } S_{n}=\frac{n}{2}[2 a+(n-1) d] \ 	ext { b) } S_{n}=\frac{a\left(1-r^{n}ight)}{1-r}\end{array} $$

UpStudy AI · Accepted Answer

Let's solve each part of the problems you've provided step by step.

---

## **Problem 3**

### **a) Calculate $$ \sum_{n=0}^{4} 3n^{2} $$**

We evaluate the sum by plugging in each value of $$ n $$ from 0 to 4:

$$
\begin{align*}
\sum_{n=0}^{4} 3n^{2} &= 3(0)^2 + 3(1)^2 + 3(2)^2 + 3(3)^2 + 3(4)^2 \
&= 0 + 3(1) + 3(4) + 3(9) + 3(16) \
&= 0 + 3 + 12 + 27 + 48 \
&= 90
\end{align*}
$$

**Answer:** $$ 90 $$

---

### **b) Calculate $$ \sum_{n=1}^{10} (90 - 3n) $$**

We can split the sum into two separate sums:

$$
\begin{align*}
\sum_{n=1}^{10} (90 - 3n) &= \sum_{n=1}^{10} 90 - \sum_{n=1}^{10} 3n \
&= 90 \times 10 - 3 \sum_{n=1}^{10} n \
&= 900 - 3 \left( \frac{10 \times 11}{2} \right) \
&= 900 - 3 \times 55 \
&= 900 - 165 \
&= 735
\end{align*}
$$

**Answer:** $$ 735 $$

---

### **c) Calculate $$ \sum_{k=5}^{25} (2k - 1) $$**

This is the sum of an arithmetic series of odd numbers from $$ k = 5 $$ to $$ k = 25 $$.

First, find the total sum from $$ k = 1 $$ to $$ k = 25 $$, then subtract the sum from $$ k = 1 $$ to $$ k = 4 $$:

$$
\begin{align*}
\sum_{k=5}^{25} (2k - 1) &= \sum_{k=1}^{25} (2k - 1) - \sum_{k=1}^{4} (2k - 1) \
&= 25^2 - 4^2 \
&= 625 - 16 \
&= 609
\end{align*}
$$

**Answer:** $$ 609 $$

---

### **d) Calculate $$ \sum_{n=1}^{13} 5\left(\frac{3}{2}\right)^{n-1} $$**

This is a geometric series with:
- First term ($$ a $$) = $$ 5 \left(\frac{3}{2}\right)^{0} = 5 $$
- Common ratio ($$ r $$) = $$ \frac{3}{2} $$
- Number of terms ($$ n $$) = 13

The sum of the first $$ n $$ terms of a geometric series is:

$$
S_n = a \frac{r^n - 1}{r - 1}
$$

Plugging in the values:

$$
\begin{align*}
S_{13} &= 5 \times \frac{\left(\frac{3}{2}\right)^{13} - 1}{\frac{3}{2} - 1} \
&= 5 \times \frac{\left(\frac{3}{2}\right)^{13} - 1}{\frac{1}{2}} \
&= 10 \left[ \left(\frac{3}{2}\right)^{13} - 1 \right]
\end{align*}
$$

**Answer:** 
$$
10 \left( \left(\frac{3}{2}\right)^{13} - 1 \right)
$$

*Alternatively, if numerical approximation is needed:*

$$
\left(\frac{3}{2}\right)^{13} \approx 159.4323
$$
$$
S_{13} \approx 10 (159.4323 - 1) = 10 \times 158.4323 = 1,584.323
$$

---

### **e) Calculate $$ \sum_{n=1}^{\infty} 3\left(\frac{1}{2}\right)^{n-1} $$**

This is an infinite geometric series with:
- First term ($$ a $$) = $$ 3 \left(\frac{1}{2}\right)^{0} = 3 $$
- Common ratio ($$ r $$) = $$ \frac{1}{2} $$

For $$ |r| < 1 $$, the sum of an infinite geometric series is:

$$
S = \frac{a}{1 - r}
$$

Plugging in the values:

$$
S = \frac{3}{1 - \frac{1}{2}} = \frac{3}{\frac{1}{2}} = 6
$$

**Answer:** $$ 6 $$

---

## **Problem 4**

### **a) Derive the formula for the sum of an arithmetic series: $$ S_{n} = \frac{n}{2}[2a + (n-1)d] $$**

**Derivation:**

Consider an arithmetic series with:
- First term ($$ a $$)
- Common difference ($$ d $$)
- Number of terms ($$ n $$)

The series can be written as:
$$
S_n = a + (a + d) + (a + 2d) + \cdots + [a + (n-1)d]
$$

Write the sum forwards and backwards:

$$
\begin{align*}
S_n &= a + (a + d) + (a + 2d) + \cdots + [a + (n-1)d] \
S_n &= [a + (n-1)d] + [a + (n-2)d] + \cdots + (a + d) + a
\end{align*}
$$

Add the two equations term by term:

$$
2S_n = [2a + (n-1)d] + [2a + (n-1)d] + \cdots + [2a + (n-1)d]
$$
$$
2S_n = n[2a + (n-1)d]
$$

Divide both sides by 2:

$$
S_n = \frac{n}{2}[2a + (n-1)d]
$$

**Final Formula:**

$$
S_n = \frac{n}{2}[2a + (n-1)d]
$$

---

### **b) Derive the formula for the sum of a geometric series: $$ S_{n} = \frac{a(1 - r^{n})}{1 - r} $$**

**Derivation:**

Consider a geometric series with:
- First term ($$ a $$)
- Common ratio ($$ r $$)
- Number of terms ($$ n $$)

The series can be written as:
$$
S_n = a + ar + ar^2 + \cdots + ar^{n-1}
$$

Multiply both sides by $$ r $$:

$$
rS_n = ar + ar^2 + ar^3 + \cdots + ar^{n}
$$

Subtract the second equation from the first:

$$
S_n - rS_n = a - ar^{n}
$$
$$
S_n(1 - r) = a(1 - r^{n})
$$

Solve for $$ S_n $$:

$$
S_n = \frac{a(1 - r^{n})}{1 - r}
$$

**Final Formula:**

$$
S_n = \frac{a(1 - r^{n})}{1 - r}
$$

---
**Problem 3 Solutions:** a) $$ \sum_{n=0}^{4} 3n^{2} = 90 $$ b) $$ \sum_{n=1}^{10} (90 - 3n) = 735 $$ c) $$ \sum_{k=5}^{25} (2k - 1) = 609 $$ d) $$ \sum_{n=1}^{13} 5\left(\frac{3}{2}\right)^{n-1} = 10 \left( \left(\frac{3}{2}\right)^{13} - 1 \right) \approx 1,584.323 $$ e) $$ \sum_{n=1}^{\infty} 3\left(\frac{1}{2}\right)^{n-1} = 6 $$ **Problem 4 Solutions:** a) $$ S_{n} = \frac{n}{2}[2a + (n-1)d] $$ b) $$ S_{n} = \frac{a(1 - r^{n})}{1 - r} $$

Calculate:

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