Certainly! Let's analyze and understand the given sequence [tex]\( a_n = \frac{n^2}{2^n} \)[/tex].
### Step-by-Step Solution:
1. Identify the Sequence Formula:
The term [tex]\( a_n \)[/tex] is defined as:
[tex]\[
a_n = \frac{n^2}{2^n}
\][/tex]
2. Understand the Behavior:
This sequence involves two components: the numerator [tex]\( n^2 \)[/tex] and the denominator [tex]\( 2^n \)[/tex].
- The numerator [tex]\( n^2 \)[/tex] grows quadratically with increasing [tex]\( n \)[/tex].
- The denominator [tex]\( 2^n \)[/tex] grows exponentially with increasing [tex]\( n \)[/tex].
3. Plugging in Specific Values:
To better understand the behavior of the sequence, let’s plug in a few specific values for [tex]\( n \)[/tex]:
- For [tex]\( n = 1 \)[/tex]:
[tex]\[
a_1 = \frac{1^2}{2^1} = \frac{1}{2} = 0.5
\][/tex]
- For [tex]\( n = 2 \)[/tex]:
[tex]\[
a_2 = \frac{2^2}{2^2} = \frac{4}{4} = 1
\][/tex]
- For [tex]\( n = 3 \)[/tex]:
[tex]\[
a_3 = \frac{3^2}{2^3} = \frac{9}{8} = 1.125
\][/tex]
- For [tex]\( n = 4 \)[/tex]:
[tex]\[
a_4 = \frac{4^2}{2^4} = \frac{16}{16} = 1
\][/tex]
- For [tex]\( n = 5 \)[/tex]:
[tex]\[
a_5 = \frac{5^2}{2^5} = \frac{25}{32} = 0.78125
\][/tex]
4. Analyze Long-term Behavior:
As [tex]\( n \)[/tex] increases, the quadratic growth of [tex]\( n^2 \)[/tex] will be outpaced by the exponential growth of [tex]\( 2^n \)[/tex]. This suggests that [tex]\( a_n \)[/tex] will decrease as [tex]\( n \)[/tex] becomes very large.
5. Summarizing the Sequence:
- Initially, the terms of the sequence may increase up to a point.
- After a certain point, the exponential growth of the denominator [tex]\( 2^n \)[/tex] will dominate, causing the overall value of [tex]\( a_n \)[/tex] to decrease.
- As [tex]\( n \)[/tex] approaches infinity, [tex]\( a_n \)[/tex] approaches 0.
6. Conclusion:
The sequence [tex]\( \frac{n^2}{2^n} \)[/tex] provides an example where initial terms might display different behavior (e.g., increase), but eventually, due to the stronger exponential growth in the denominator, the terms of the sequence decrease and approach zero as [tex]\( n \)[/tex] becomes very large.
Thus, we have a clear understanding of the sequence [tex]\( a_n = \frac{n^2}{2^n} \)[/tex] both in terms of specific values and its general behavior.
