Question 1
Solution
(a) Show that $ \sum_{r=1}^n (3r + 2) = \frac{n}{2}(3n + 7) $
We start by expanding the sum:
$ \sum_{r=1}^n (3r + 2) = \sum_{r=1}^n 3r + \sum_{r=1}^n 2 $
Step 1: Evaluate $ \sum_{r=1}^n 3r $
Factor out the constant $3$:
$ \sum_{r=1}^n 3r = 3 \sum_{r=1}^n r $
The sum of the first $n$ integers is given by the formula:
$ \sum_{r=1}^n r = \frac{n(n+1)}{2} $
Thus:
$ \sum_{r=1}^n 3r = 3 \cdot \frac{n(n+1)}{2} = \frac{3n(n+1)}{2} $
Step 2: Evaluate $ \sum_{r=1}^n 2 $
Since $2$ is a constant:
$ \sum_{r=1}^n 2 = 2n $
Step 3: Combine the two sums
Now we can combine both results:
$ \sum_{r=1}^n (3r + 2) = \frac{3n(n+1)}{2} + 2n $
Step 4: Simplify the expression
To simplify, express $2n$ with a denominator of 2:
$ \sum_{r=1}^n (3r + 2) = \frac{3n(n+1)}{2} + \frac{4n}{2} $
Now combine the two terms:
$ \sum_{r=1}^n (3r + 2) = \frac{3n^2 + 3n + 4n}{2} = \frac{3n^2 + 7n}{2} $
Finally, factor the result:
$ \sum_{r=1}^n (3r + 2) = \frac{n}{2}(3n + 7) $
(b) Hence, or otherwise, evaluate $ \sum_{r=10}^{40} (3r + 2) $
We can use the result from part (a) to calculate the sum from $r = 10$ to $r = 40$:
$ \sum_{r=10}^{40} (3r + 2) = \sum_{r=1}^{40} (3r + 2) - \sum_{r=1}^{9} (3r + 2) $
Step 1: Evaluate $ \sum_{r=1}^{40} (3r + 2) $
Using the formula from part (a):
$ \sum_{r=1}^{40} (3r + 2) = \frac{40}{2}(3 \cdot 40 + 7) = 20 \cdot (120 + 7) = 20 \cdot 127 = 2540 $
Step 2: Evaluate $ \sum_{r=1}^{9} (3r + 2) $
Using the formula from part (a):
$ \sum_{r=1}^{9} (3r + 2) = \frac{9}{2}(3 \cdot 9 + 7) = \frac{9}{2} \cdot (27 + 7) = \frac{9}{2} \cdot 34 = 153 $
Step 3: Calculate the desired sum
Now subtract the two sums:
$ \sum_{r=10}^{40} (3r + 2) = 2540 - 153 = 2387 $
Thus, the sum is:
$ \sum_{r=10}^{40} (3r + 2) = 2387 $
Question 2
Solution
Problem:
$ \begin{aligned} y &= (\sin 2x)\sqrt{3 + 2x} \\ \text{Show that } \frac{dy}{dx} &= \frac{\sin 2x + (A + Bx)\cos 2x}{\sqrt{3 + 2x}} \text{ where } A \text{ and } B \text{ are integers to be found.} \end{aligned} $
Solution:
We start by applying the product rule to the expression for $y$:
$ y = (\sin 2x)\sqrt{3 + 2x} $
Step 1: Apply the product rule.
The product rule states that:
$ \frac{dy}{dx} = \frac{d}{dx}(\sin 2x)\cdot\sqrt{3 + 2x} + \sin 2x \cdot \frac{d}{dx}(\sqrt{3 + 2x}) $
Step 2: Differentiate each part.
First term: Differentiate $ \sin 2x $.
Using the chain rule:
$ \frac{d}{dx}(\sin 2x) = 2\cos 2x $
Second term: Differentiate $ \sqrt{3 + 2x} $.
Using the chain rule:
$ \frac{d}{dx}(\sqrt{3 + 2x}) = \frac{1}{2\sqrt{3 + 2x}} \cdot \frac{d}{dx}(3 + 2x) = \frac{1}{2\sqrt{3 + 2x}} \cdot 2 = \frac{1}{\sqrt{3 + 2x}} $
Step 3: Substitute the derivatives into the product rule.
$ \frac{dy}{dx} = (2\cos 2x)\cdot\sqrt{3 + 2x} + (\sin 2x)\cdot \frac{1}{\sqrt{3 + 2x}} $
This simplifies to:
$ \frac{dy}{dx} = 2\cos 2x \cdot \sqrt{3 + 2x} + \frac{\sin 2x}{\sqrt{3 + 2x}} $
Step 4: Factor out $ \frac{1}{\sqrt{3 + 2x}} $.
$ \frac{dy}{dx} = \frac{1}{\sqrt{3 + 2x}} \left( 2\cos 2x \cdot (3 + 2x) + \sin 2x \right) $
Step 5: Expand the expression inside the parentheses.
$ 2\cos 2x \cdot (3 + 2x) = 6\cos 2x + 4x\cos 2x $
Thus:
$ \frac{dy}{dx} = \frac{1}{\sqrt{3 + 2x}} \left( 6\cos 2x + 4x\cos 2x + \sin 2x \right) $
Step 6: Match the desired form.
Compare with:
$ \frac{\sin 2x + (A + Bx)\cos 2x}{\sqrt{3 + 2x}} $
We see that:
- $\sin 2x$ matches directly
- $(A + Bx)\cos 2x = 6\cos 2x + 4x\cos 2x$
Therefore:
$ A = 6 \quad \text{and} \quad B = 4 $
Final Answer:
$ \frac{dy}{dx} = \frac{\sin 2x + (6 + 4x)\cos 2x}{\sqrt{3 + 2x}} $
where $A = 6$ and $B = 4$.
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