Calculate The Sum Of 1/a_n: A Detailed Solution

by Admin 48 views
Calculate the Sum of 1/a_n: A Detailed Solution

Hey guys! Let's dive into an exciting mathematical problem today. We're going to tackle the challenge of finding the sum of 1an\frac{1}{a_n} given a specific definition for ana_n. This problem involves some algebraic manipulation and a keen eye for simplification. So, buckle up and let's get started!

Understanding the Problem

First, let's clearly state the problem. We are given that an=1+(1+1n)2+1+(1−1n)2a_n = \sqrt{1+\left(1+\frac{1}{n}\right)^2} + \sqrt{1+\left(1-\frac{1}{n}\right)^2}. Our mission, should we choose to accept it (and we do!), is to find the value of the sum:

∑n=1201an=1a1+1a2+1a3+...+1a20\sum_{n=1}^{20} \frac{1}{a_n} = \frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} + ... + \frac{1}{a_{20}}

This looks intimidating at first glance, but don't worry! We'll break it down step by step. The key here is to simplify the expression for ana_n and then find a way to make the summation manageable. Remember, in math, complex problems often have elegant solutions hiding beneath the surface. Our job is to uncover that elegance. We'll use algebraic techniques and a bit of mathematical intuition to navigate through this. Let's start by simplifying the expression inside the square roots.

Simplifying the Expression for a_n

Our first task is to simplify the expression for ana_n. Let's start by expanding the terms inside the square roots. We have:

an=1+(1+1n)2+1+(1−1n)2a_n = \sqrt{1+\left(1+\frac{1}{n}\right)^2} + \sqrt{1+\left(1-\frac{1}{n}\right)^2}

Let's expand the squares:

(1+1n)2=1+2n+1n2\left(1+\frac{1}{n}\right)^2 = 1 + \frac{2}{n} + \frac{1}{n^2}

(1−1n)2=1−2n+1n2\left(1-\frac{1}{n}\right)^2 = 1 - \frac{2}{n} + \frac{1}{n^2}

Now, substitute these back into the expression for ana_n:

an=1+1+2n+1n2+1+1−2n+1n2a_n = \sqrt{1 + 1 + \frac{2}{n} + \frac{1}{n^2}} + \sqrt{1 + 1 - \frac{2}{n} + \frac{1}{n^2}}

Which simplifies to:

an=2+2n+1n2+2−2n+1n2a_n = \sqrt{2 + \frac{2}{n} + \frac{1}{n^2}} + \sqrt{2 - \frac{2}{n} + \frac{1}{n^2}}

This looks a bit cleaner, but we can still do better. Let's try to rewrite the expressions inside the square roots as perfect squares. Notice that we can rewrite the terms as:

2+2n+1n2=(1+1n)2+12 + \frac{2}{n} + \frac{1}{n^2} = \left(1 + \frac{1}{n}\right)^2 + 1

and

2−2n+1n2=(1−1n)2+12 - \frac{2}{n} + \frac{1}{n^2} = \left(1 - \frac{1}{n}\right)^2 + 1

However, this doesn't directly lead to a simplification under the square root. Instead, let's try factoring out 1n2\frac{1}{n^2} from the terms inside the square roots:

an=1n2(2n2+2n+1)+1n2(2n2−2n+1)a_n = \sqrt{\frac{1}{n^2}(2n^2 + 2n + 1)} + \sqrt{\frac{1}{n^2}(2n^2 - 2n + 1)}

This gives us:

an=1n2n2+2n+1+1n2n2−2n+1a_n = \frac{1}{n}\sqrt{2n^2 + 2n + 1} + \frac{1}{n}\sqrt{2n^2 - 2n + 1}

Now, we have a slightly more manageable form for ana_n. Next, we'll focus on finding an expression for 1an\frac{1}{a_n}.

Finding 1/a_n

Now that we have a simplified expression for ana_n, let's find 1an\frac{1}{a_n}. Remember, this is a crucial step because we need to evaluate the sum of 1an\frac{1}{a_n}.

We have:

an=1n2n2+2n+1+1n2n2−2n+1a_n = \frac{1}{n}\sqrt{2n^2 + 2n + 1} + \frac{1}{n}\sqrt{2n^2 - 2n + 1}

So,

1an=11n2n2+2n+1+1n2n2−2n+1\frac{1}{a_n} = \frac{1}{\frac{1}{n}\sqrt{2n^2 + 2n + 1} + \frac{1}{n}\sqrt{2n^2 - 2n + 1}}

1an=n2n2+2n+1+2n2−2n+1\frac{1}{a_n} = \frac{n}{\sqrt{2n^2 + 2n + 1} + \sqrt{2n^2 - 2n + 1}}

This looks complicated, but we can simplify it further by rationalizing the denominator. We'll multiply the numerator and denominator by the conjugate of the denominator:

1an=n2n2+2n+1+2n2−2n+1⋅2n2+2n+1−2n2−2n+12n2+2n+1−2n2−2n+1\frac{1}{a_n} = \frac{n}{\sqrt{2n^2 + 2n + 1} + \sqrt{2n^2 - 2n + 1}} \cdot \frac{\sqrt{2n^2 + 2n + 1} - \sqrt{2n^2 - 2n + 1}}{\sqrt{2n^2 + 2n + 1} - \sqrt{2n^2 - 2n + 1}}

Multiplying this out, we get:

1an=n(2n2+2n+1−2n2−2n+1)(2n2+2n+1)−(2n2−2n+1)\frac{1}{a_n} = \frac{n(\sqrt{2n^2 + 2n + 1} - \sqrt{2n^2 - 2n + 1})}{(2n^2 + 2n + 1) - (2n^2 - 2n + 1)}

1an=n(2n2+2n+1−2n2−2n+1)4n\frac{1}{a_n} = \frac{n(\sqrt{2n^2 + 2n + 1} - \sqrt{2n^2 - 2n + 1})}{4n}

Simplifying, we get:

1an=2n2+2n+1−2n2−2n+14\frac{1}{a_n} = \frac{\sqrt{2n^2 + 2n + 1} - \sqrt{2n^2 - 2n + 1}}{4}

Now, this is a much cleaner expression for 1an\frac{1}{a_n}! The next step is to evaluate the sum using this simplified form. This is where we'll see if we can find a pattern that allows us to simplify the summation.

Evaluating the Sum

Now that we have a simplified expression for 1an\frac{1}{a_n}, we can evaluate the sum:

∑n=1201an=∑n=1202n2+2n+1−2n2−2n+14\sum_{n=1}^{20} \frac{1}{a_n} = \sum_{n=1}^{20} \frac{\sqrt{2n^2 + 2n + 1} - \sqrt{2n^2 - 2n + 1}}{4}

We can rewrite this as:

14∑n=120(2n2+2n+1−2n2−2n+1)\frac{1}{4} \sum_{n=1}^{20} (\sqrt{2n^2 + 2n + 1} - \sqrt{2n^2 - 2n + 1})

This sum looks like a telescoping series, which is a series where intermediate terms cancel out, leaving only the first and last terms. To see this more clearly, let's write out a few terms of the sum:

For n = 1: 2(1)2+2(1)+1−2(1)2−2(1)+1=5−1\sqrt{2(1)^2 + 2(1) + 1} - \sqrt{2(1)^2 - 2(1) + 1} = \sqrt{5} - \sqrt{1}

For n = 2: 2(2)2+2(2)+1−2(2)2−2(2)+1=13−5\sqrt{2(2)^2 + 2(2) + 1} - \sqrt{2(2)^2 - 2(2) + 1} = \sqrt{13} - \sqrt{5}

For n = 3: 2(3)2+2(3)+1−2(3)2−2(3)+1=25−13\sqrt{2(3)^2 + 2(3) + 1} - \sqrt{2(3)^2 - 2(3) + 1} = \sqrt{25} - \sqrt{13}

Notice the pattern? The 5\sqrt{5} in the first term cancels with the −5-\sqrt{5} in the second term, and the 13\sqrt{13} in the second term cancels with the −13-\sqrt{13} in the third term. This pattern continues throughout the series. So, let's write out the general form:

The sum can be written as:

14[(5−1)+(13−5)+(25−13)+...+(2(20)2+2(20)+1−2(20)2−2(20)+1)]\frac{1}{4}[(\sqrt{5} - \sqrt{1}) + (\sqrt{13} - \sqrt{5}) + (\sqrt{25} - \sqrt{13}) + ... + (\sqrt{2(20)^2 + 2(20) + 1} - \sqrt{2(20)^2 - 2(20) + 1})]

The intermediate terms will cancel out, leaving us with the first and last terms. The last term corresponds to n = 20:

2(20)2+2(20)+1−2(20)2−2(20)+1=841−761\sqrt{2(20)^2 + 2(20) + 1} - \sqrt{2(20)^2 - 2(20) + 1} = \sqrt{841} - \sqrt{761}

So, the sum telescopes to:

14(2(20)2+2(20)+1−2(1)2−2(1)+1)\frac{1}{4}(\sqrt{2(20)^2 + 2(20) + 1} - \sqrt{2(1)^2 - 2(1) + 1})

14(841−1)\frac{1}{4}(\sqrt{841} - \sqrt{1})

Now, we know that 841=29\sqrt{841} = 29 and 1=1\sqrt{1} = 1, so the expression simplifies to:

14(29−1)=14(28)=7\frac{1}{4}(29 - 1) = \frac{1}{4}(28) = 7

Therefore, the value of the sum is 7.

Conclusion

We've successfully found the value of the sum ∑n=1201an\sum_{n=1}^{20} \frac{1}{a_n}! We started with a seemingly complex expression for ana_n, but through careful simplification and the clever use of rationalization, we were able to find a manageable form for 1an\frac{1}{a_n}. The key to solving this problem was recognizing the telescoping nature of the series. By writing out a few terms, we saw the pattern of cancellation, which allowed us to simplify the sum and arrive at the final answer of 7.

So, the next time you encounter a daunting mathematical problem, remember to break it down into smaller steps, look for patterns, and don't be afraid to get your hands dirty with algebraic manipulations. You might just uncover an elegant solution hiding beneath the complexity!