Let's break down each sum and solve them one by one.

Part a)

The sum is given as:

$\sum_{i=0}^{3} \frac{0.6(i^3 - 2)}{i + 1}$

We need to calculate this for $i = 0$, $i = 1$, $i = 2$, and $i = 3$.

1. For $i = 0$: $\frac{0.6(0^3 - 2)}{0 + 1} = \frac{0.6(-2)}{1} = -1.2$

2. For $i = 1$: $\frac{0.6(1^3 - 2)}{1 + 1} = \frac{0.6(1 - 2)}{2} = \frac{0.6(-1)}{2} = -0.3$

3. For $i = 2$: $\frac{0.6(2^3 - 2)}{2 + 1} = \frac{0.6(8 - 2)}{3} = \frac{0.6(6)}{3} = 1.2$

4. For $i = 3$: $\frac{0.6(3^3 - 2)}{3 + 1} = \frac{0.6(27 - 2)}{4} = \frac{0.6(25)}{4} = 3.75$

Now, adding these values together: $-1.2 + (-0.3) + 1.2 + 3.75 = 3.45$

So, the sum for part a is: $\boxed{3.45}$

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Part b)

The sum is given as:

$\sum_{n=0}^{3} 48 \cdot \left(-\frac{1}{4}\right)^n$

We need to compute this for $n = 0$, $n = 1$, $n = 2$, and $n = 3$.

1. For $n = 0$: $48 \cdot \left(-\frac{1}{4}\right)^0 = 48 \cdot 1 = 48$

2. For $n = 1$: $48 \cdot \left(-\frac{1}{4}\right)^1 = 48 \cdot \left(-\frac{1}{4}\right) = -12$

3. For $n = 2$: $48 \cdot \left(-\frac{1}{4}\right)^2 = 48 \cdot \frac{1}{16} = 3$

4. For $n = 3$: $48 \cdot \left(-\frac{1}{4}\right)^3 = 48 \cdot \left(-\frac{1}{64}\right) = -0.75$

Now, adding these values together: $48 + (-12) + 3 + (-0.75) = 38.25$

So, the sum for part b is: $\boxed{38.25}$

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Final Answers:

• Part a: 3.45
• Part b: 38.25

Would you like any further explanations on how I got these results or any additional steps? Here are some related questions for deeper understanding:

1. How do you calculate the general formula for summation of polynomials?
2. What are the applications of geometric series in real-world problems?
3. How do you handle negative exponents in summation problems?
4. What happens to the value of a sum as the upper limit increases in summations like part b?
5. Can we use integration to approximate the sum of polynomial terms?

Tip: Always double-check the limits of summation and ensure each term is computed correctly in sequence!