Smallest Term In Sequence: Solve 5 Tricky Examples

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Hey guys! Ever wondered how to pinpoint the smallest value in a sequence? It's a common problem in algebra, and this guide will walk you through finding the smallest term in various sequences. We'll break down five different examples, so you'll be a pro in no time. Let's dive in!

Understanding Sequences

Before we jump into solving, let's quickly recap what a sequence is. A sequence is simply an ordered list of numbers. Each number in the sequence is called a term, and we often denote the n{n}-th term as xn{x_n}. Our mission is to find the minimum value among all these terms. This involves understanding the behavior of the sequence as n{n} changes, sometimes increasing and sometimes decreasing. Understanding sequences is the fundamental step in identifying patterns and minimum values. A sequence is essentially a function whose domain is the set of natural numbers. Therefore, to find the smallest term, we need to analyze the function's behavior.

Why is This Important?

Knowing how to find the smallest term isn't just a math exercise. It has practical applications in various fields, such as computer science (analyzing algorithm efficiency), economics (modeling cost minimization), and physics (studying energy states). So, understanding this concept is super useful! Consider situations where you need to optimize processes or minimize costs; this skill becomes invaluable. For example, in algorithm design, finding the minimum steps to complete a task can significantly improve performance. In economics, businesses often aim to minimize production costs while maximizing profits. In physics, understanding minimum energy states helps in predicting the stability of systems. Thus, mastering the technique to find the smallest term provides a versatile toolset for problem-solving across multiple domains.

General Strategies

To find the smallest term in a sequence, we'll use a combination of algebraic manipulation, calculus concepts (if applicable), and logical reasoning. Here are some common strategies:

  • Analyzing the Ratio xn+1xn{\frac{x_{n+1}}{x_n}}: If this ratio is less than 1, the sequence is decreasing; if it's greater than 1, the sequence is increasing. When the ratio changes from less than 1 to greater than 1, we've likely found the minimum term.
  • Finding the Derivative (for continuous functions): If we can treat the sequence as a continuous function, we can find its derivative, set it to zero, and solve for critical points. These points can indicate local minima.
  • Completing the Square (for quadratic sequences): This helps rewrite the quadratic expression in a form that reveals the vertex, which corresponds to the minimum or maximum value.
  • Direct Substitution: Sometimes, the easiest way is to simply calculate the first few terms and look for a pattern.

Example 1: xn=n32n{x_n = \frac{n^3}{2^n}}

Let's start with our first sequence: xn=n32n{x_n = \frac{n^3}{2^n}}. To find the smallest term here, we'll analyze the ratio xn+1xn{\frac{x_{n+1}}{x_n}}. This approach helps us understand when the sequence transitions from decreasing to increasing, which will pinpoint our minimum.

Analyzing the Ratio

We need to calculate xn+1xn{\frac{x_{n+1}}{x_n}}:

xn+1xn=(n+1)32n+1n32n=(n+1)32n+1⋅2nn3=(n+1)32n3{ \frac{x_{n+1}}{x_n} = \frac{\frac{(n+1)^3}{2^{n+1}}}{\frac{n^3}{2^n}} = \frac{(n+1)^3}{2^{n+1}} \cdot \frac{2^n}{n^3} = \frac{(n+1)^3}{2n^3} }

Now, we want to find when this ratio is less than 1 (decreasing sequence) and when it's greater than 1 (increasing sequence):

(n+1)32n3<1{ \frac{(n+1)^3}{2n^3} < 1 }

(n+1)3<2n3{ (n+1)^3 < 2n^3 }

Expanding and simplifying:

n3+3n2+3n+1<2n3{ n^3 + 3n^2 + 3n + 1 < 2n^3 }

0<n3−3n2−3n−1{ 0 < n^3 - 3n^2 - 3n - 1 }

This inequality is a bit tricky to solve analytically, but we can analyze it numerically. We can test different values of n to see when the inequality holds true.

Numerical Analysis

Let's try some values for n{n}:

  • For n=1{n = 1}: 1−3−3−1=−6<0{1 - 3 - 3 - 1 = -6 < 0}
  • For n=2{n = 2}: 8−12−6−1=−11<0{8 - 12 - 6 - 1 = -11 < 0}
  • For n=3{n = 3}: 27−27−9−1=−10<0{27 - 27 - 9 - 1 = -10 < 0}
  • For n=4{n = 4}: 64−48−12−1=3>0{64 - 48 - 12 - 1 = 3 > 0}

So, the inequality holds for n≥4{n \geq 4}. This means that the sequence decreases until n=4{n = 4} and then starts increasing. Therefore, the smallest term is likely around n=3{n = 3} or n=4{n = 4}. It's important to test a few values around the critical point to ensure we've found the true minimum.

Calculating Terms

Let's calculate x3{x_3}, x4{x_4}, and x5{x_5}:

x3=3323=278=3.375{ x_3 = \frac{3^3}{2^3} = \frac{27}{8} = 3.375 }

x4=4324=6416=4{ x_4 = \frac{4^3}{2^4} = \frac{64}{16} = 4 }

x5=5325=12532=3.90625{ x_5 = \frac{5^3}{2^5} = \frac{125}{32} = 3.90625 }

Comparing these values, we see that the smallest term is x3=3.375{x_3 = 3.375}.

Final Answer: The smallest term in the sequence xn=n32n{x_n = \frac{n^3}{2^n}} is x3=278{x_3 = \frac{27}{8}}.

Example 2: xn=n100+n{x_n = \frac{\sqrt{n}}{100 + n}}

Now let's tackle the second sequence: xn=n100+n{x_n = \frac{\sqrt{n}}{100 + n}}. This one looks a bit different, but we can still use the ratio approach or consider treating it as a continuous function. For this example, we'll use the ratio test again to find the smallest term.

Analyzing the Ratio

Let's find xn+1xn{\frac{x_{n+1}}{x_n}}:

xn+1xn=n+1100+(n+1)n100+n=n+1101+n⋅100+nn{ \frac{x_{n+1}}{x_n} = \frac{\frac{\sqrt{n+1}}{100 + (n+1)}}{\frac{\sqrt{n}}{100 + n}} = \frac{\sqrt{n+1}}{101 + n} \cdot \frac{100 + n}{\sqrt{n}} }

xn+1xn=n+1(100+n)n(101+n){ \frac{x_{n+1}}{x_n} = \frac{\sqrt{n+1}(100 + n)}{\sqrt{n}(101 + n)} }

To determine when the sequence is decreasing, we need to find when this ratio is less than 1:

n+1(100+n)n(101+n)<1{ \frac{\sqrt{n+1}(100 + n)}{\sqrt{n}(101 + n)} < 1 }

n+1(100+n)<n(101+n){ \sqrt{n+1}(100 + n) < \sqrt{n}(101 + n) }

Squaring both sides to get rid of the square roots:

(n+1)(100+n)2<n(101+n)2{ (n+1)(100 + n)^2 < n(101 + n)^2 }

Expanding and simplifying:

(n+1)(10000+200n+n2)<n(10201+202n+n2){ (n+1)(10000 + 200n + n^2) < n(10201 + 202n + n^2) }

10000n+200n2+n3+10000+200n+n2<10201n+202n2+n3{ 10000n + 200n^2 + n^3 + 10000 + 200n + n^2 < 10201n + 202n^2 + n^3 }

10000n+201n2+10000+200n<10201n+202n2{ 10000n + 201n^2 + 10000 + 200n < 10201n + 202n^2 }

0<n2+n−10000{ 0 < n^2 + n - 10000 }

Solving the Inequality

We have a quadratic inequality. To solve n2+n−10000>0{n^2 + n - 10000 > 0}, we first find the roots of the corresponding quadratic equation n2+n−10000=0{n^2 + n - 10000 = 0} using the quadratic formula:

n=−b±b2−4ac2a=−1±12−4(1)(−10000)2(1){ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1^2 - 4(1)(-10000)}}{2(1)} }

n=−1±1+400002=−1±400012{ n = \frac{-1 \pm \sqrt{1 + 40000}}{2} = \frac{-1 \pm \sqrt{40001}}{2} }

The positive root is approximately:

n≈−1+200.00252≈99.5{ n \approx \frac{-1 + 200.0025}{2} \approx 99.5 }

Since the parabola opens upwards, the inequality n2+n−10000>0{n^2 + n - 10000 > 0} holds for n>99.5{n > 99.5}. This suggests that the sequence decreases until around n=99{n = 99} or n=100{n = 100} and then starts increasing. Again, testing values around the critical point is essential.

Calculating Terms

Let's calculate x99{x_{99}}, x100{x_{100}}, and x101{x_{101}}:

x99=99100+99≈9.95199≈0.04997{ x_{99} = \frac{\sqrt{99}}{100 + 99} \approx \frac{9.95}{199} \approx 0.04997 }

x100=100100+100=10200=0.05{ x_{100} = \frac{\sqrt{100}}{100 + 100} = \frac{10}{200} = 0.05 }

x101=101100+101≈10.05201≈0.05002{ x_{101} = \frac{\sqrt{101}}{100 + 101} \approx \frac{10.05}{201} \approx 0.05002 }

It appears that the smallest term is around x99{x_{99}}.

Final Answer: The smallest term in the sequence xn=n100+n{x_n = \frac{\sqrt{n}}{100 + n}} is approximately x99≈0.04997{x_{99} \approx 0.04997}.

Example 3: xn=1000nn!{x_n = \frac{1000^n}{n!}}

Let's move on to the third sequence: xn=1000nn!{x_n = \frac{1000^n}{n!}}. This one involves factorials, so the ratio test will be particularly useful. We'll use the ratio xn+1xn{\frac{x_{n+1}}{x_n}} to find the smallest term.

Analyzing the Ratio

We calculate xn+1xn{\frac{x_{n+1}}{x_n}}:

xn+1xn=1000n+1(n+1)!1000nn!=1000n+1(n+1)!⋅n!1000n{ \frac{x_{n+1}}{x_n} = \frac{\frac{1000^{n+1}}{(n+1)!}}{\frac{1000^n}{n!}} = \frac{1000^{n+1}}{(n+1)!} \cdot \frac{n!}{1000^n} }

xn+1xn=1000n+11000n⋅n!(n+1)!=1000⋅1n+1=1000n+1{ \frac{x_{n+1}}{x_n} = \frac{1000^{n+1}}{1000^n} \cdot \frac{n!}{(n+1)!} = 1000 \cdot \frac{1}{n+1} = \frac{1000}{n+1} }

We want to find when this ratio is less than 1 (decreasing sequence):

1000n+1<1{ \frac{1000}{n+1} < 1 }

1000<n+1{ 1000 < n+1 }

999<n{ 999 < n }

So, the sequence decreases when n>999{n > 999} and increases when n<999{n < 999}. This means the smallest term will be around n=999{n = 999} or n=1000{n = 1000}. It's crucial to calculate terms around the transition point.

Calculating Terms

The sequence decreases until n=999{n = 999}, and then it starts to increase. Therefore, the smallest term is either x999{x_{999}} or x1000{x_{1000}}. Since x1000x999=1{\frac{x_{1000}}{x_{999}} = 1}, we have x999=x1000{x_{999} = x_{1000}}.

Final Answer: The smallest terms in the sequence xn=1000nn!{x_n = \frac{1000^n}{n!}} are x999{x_{999}} and x1000{x_{1000}}.

Example 4: xn=n2−9n−100{x_n = n^2 - 9n - 100}

Let's consider the fourth sequence: xn=n2−9n−100{x_n = n^2 - 9n - 100}. This is a quadratic sequence, and we can find the smallest term by completing the square or finding the vertex of the parabola.

Completing the Square

We complete the square:

xn=n2−9n−100=(n2−9n)−100{ x_n = n^2 - 9n - 100 = (n^2 - 9n) - 100 }

xn=(n2−9n+(92)2)−(92)2−100{ x_n = \left(n^2 - 9n + \left(\frac{9}{2}\right)^2\right) - \left(\frac{9}{2}\right)^2 - 100 }

xn=(n−92)2−814−100{ x_n = \left(n - \frac{9}{2}\right)^2 - \frac{81}{4} - 100 }

xn=(n−92)2−814−4004{ x_n = \left(n - \frac{9}{2}\right)^2 - \frac{81}{4} - \frac{400}{4} }

xn=(n−92)2−4814{ x_n = \left(n - \frac{9}{2}\right)^2 - \frac{481}{4} }

The vertex of the parabola is at n=92=4.5{n = \frac{9}{2} = 4.5}. Since n{n} must be an integer, we check the integers closest to 4.5, which are 4 and 5.

Calculating Terms

Let's calculate x4{x_4} and x5{x_5}:

x4=42−9(4)−100=16−36−100=−120{ x_4 = 4^2 - 9(4) - 100 = 16 - 36 - 100 = -120 }

x5=52−9(5)−100=25−45−100=−120{ x_5 = 5^2 - 9(5) - 100 = 25 - 45 - 100 = -120 }

Both values are the same, so we have two smallest terms.

Final Answer: The smallest terms in the sequence xn=n2−9n−100{x_n = n^2 - 9n - 100} are x4=−120{x_4 = -120} and x5=−120{x_5 = -120}.

Example 5: xn=n+100n{x_n = n + \frac{100}{n}}

Finally, let's look at the fifth sequence: xn=n+100n{x_n = n + \frac{100}{n}}. For this sequence, we can also consider it as a continuous function and use calculus, or we can analyze the behavior of the terms directly to find the smallest term.

Analyzing the Terms

Let's calculate the first few terms:

  • x1=1+1001=101{x_1 = 1 + \frac{100}{1} = 101}
  • x2=2+1002=52{x_2 = 2 + \frac{100}{2} = 52}
  • x5=5+1005=25{x_5 = 5 + \frac{100}{5} = 25}
  • x10=10+10010=20{x_{10} = 10 + \frac{100}{10} = 20}
  • x11=11+10011≈11+9.09=20.09{x_{11} = 11 + \frac{100}{11} \approx 11 + 9.09 = 20.09}

It looks like the sequence is decreasing for a while and then starts increasing. Let's try x9{x_9} and x10{x_{10}} to confirm this pattern.

x9=9+1009≈9+11.11=20.11{ x_9 = 9 + \frac{100}{9} \approx 9 + 11.11 = 20.11 }

x10=10+10010=20{ x_{10} = 10 + \frac{100}{10} = 20 }

x11=11+10011≈11+9.09=20.09{ x_{11} = 11 + \frac{100}{11} \approx 11 + 9.09 = 20.09 }

Identifying the Minimum

From these calculations, we can see that the smallest term is x10=20{x_{10} = 20}.

Final Answer: The smallest term in the sequence xn=n+100n{x_n = n + \frac{100}{n}} is x10=20{x_{10} = 20}.

Conclusion

Alright, guys! We've walked through five different examples of finding the smallest term in a sequence. We used various techniques, including analyzing ratios, completing the square, and direct calculation. Remember, the key is to understand the behavior of the sequence and choose the right approach. Keep practicing, and you'll become a sequence-solving superstar! Understanding how sequences behave is crucial in numerous mathematical and real-world scenarios, so the effort you invest in mastering these techniques is well worth it. Whether it's optimizing algorithms, minimizing costs, or solving theoretical problems, the ability to analyze sequences effectively will undoubtedly prove to be a valuable asset in your problem-solving toolkit. So, keep exploring, keep learning, and most importantly, keep having fun with math!