Sequences & Series
Sum of Infinite Series
Grade 11
Question:
<p>314. For \(d \in \left(0, \frac{\pi}{2}\right)\), if \(x = \sum_{n=0}^{\infty} \cos^{2n}\theta\), \(y = \sum_{n=0}^{\infty} \sin^{2n}\theta\), \(z = \sum_{n=0}^{\infty} \cos^{2n}\theta \cdot \sin^{2n}\theta\), then</p>
<p>(a) \(xyz = xz + y\)</p>
<p>(b) \(xyz = xy + z\)</p>
<p>(c) \(xyz = x + y + z\)</p>
<p>(d) \(xyz = yz + x\)</p>
Step-by-Step Solution
Key Concept: Recognize these as geometric series with ratios cos²θ, sin²θ, and cos²θ·sin²θ respectively. Use the constraint that x, y, z satisfy a key algebraic relation derived from xy = x + y + z after summing.
Step 1: Express x, y, and z as sums of geometric series.
Given that $\theta \in \left(0, \frac{\pi}{2}\right)$, we know that $\cos^2\theta \in (0, 1)$ and $\sin^2\theta \in (0, 1)$.
The given expressions for $x$, $y$, and $z$ are infinite geometric series with common ratios less than 1.
The sum of an infinite geometric series $S = \sum_{n=0}^{\infty} ar^n = \frac{a}{1-r}$ for $|r| < 1$. In our case, $a=1$ for all series.
For $x$:
$$x = \sum_{n=0}^{\infty} \cos^{2n}\theta = \sum_{n=0}^{\infty} (\cos^2\theta)^n = \frac{1}{1-\cos^2\theta}$$
Using the identity $\sin^2\theta + \cos^2\theta = 1$, we get $1-\cos^2\theta = \sin^2\theta$.
$$x = \frac{1}{\sin^2\theta}$$
For $y$:
$$y = \sum_{n=0}^{\infty} \sin^{2n}\theta = \sum_{n=0}^{\infty} (\sin^2\theta)^n = \frac{1}{1-\sin^2\theta}$$
Using the identity $\sin^2\theta + \cos^2\theta = 1$, we get $1-\sin^2\theta = \cos^2\theta$.
$$y = \frac{1}{\cos^2\theta}$$
For $z$:
$$z = \sum_{n=0}^{\infty} \cos^{2n}\theta \cdot \sin^{2n}\theta = \sum_{n=0}^{\infty} (\cos^2\theta \cdot \sin^2\theta)^n = \frac{1}{1-\cos^2\theta \cdot \sin^2\theta}$$
Step 2: Establish a relationship between x and y.
From Step 1, we have $x = \frac{1}{\sin^2\theta}$ and $y = \frac{1}{\cos^2\theta}$.
This implies $\sin^2\theta = \frac{1}{x}$ and $\cos^2\theta = \frac{1}{y}$.
Using the fundamental trigonometric identity $\sin^2\theta + \cos^2\theta = 1$:
$$\frac{1}{x} + \frac{1}{y} = 1$$
Combining the terms on the left side:
$$\frac{y+x}{xy} = 1$$
$$x+y = xy$$
Step 3: Establish a relationship between z and the product xy.
From Step 1, we have $z = \frac{1}{1-\cos^2\theta \cdot \sin^2\theta}$.
Substitute the expressions for $\sin^2\theta$ and $\cos^2\theta$ from Step 2:
$$z = \frac{1}{1-\left(\frac{1}{y}\right)\left(\frac{1}{x}\right)}$$
$$z = \frac{1}{1-\frac{1}{xy}}$$
To simplify the denominator, find a common denominator:
$$z = \frac{1}{\frac{xy-1}{xy}}$$
$$z = \frac{xy}{xy-1}$$
Step 4: Rearrange the relationship to match one of the given options.
We have the equation $z = \frac{xy}{xy-1}$.
Multiply both sides by $(xy-1)$:
$$z(xy-1) = xy$$
Distribute $z$ on the left side:
$$xyz - z = xy$$
Rearrange the terms to isolate $xyz$:
$$xyz = xy + z$$
This equation matches Option (b).
The final answer is $\boxed{xyz = xy + z}$.
Correct Answer: B