What is the sixth term of the sequence defined by a_n = 2^n/n^2, rounded to the nearest tenth?

What is the sixth term of the sequence defined by a_n = 2^n/n^2, rounded to the nearest tenth?

• A. 1.6
• B. 1.8
• C. 2.0
• D. 3.2

Answer: (B)

To find the sixth term of the sequence defined by \( a_n = \frac{2^n}{n^2} \), substitute \( n = 6 \) into the expression. Calculating \( a_6 \): \[ a_6 = \frac{2^6}{6^2} \] First, calculate \( 2^6 \): \[ 2^6 = 64 \] Next, compute \( 6^2 \): \[ 6^2 = 36 \] Now substitute these values back into the expression for \( a_6 \): \[ a_6 = \frac{64}{36} \] To simplify \( \frac{64}{36} \), both the numerator and the denominator can be divided by 4: \[ a_6 = \frac{16}{9} \] Now, we can convert \( \frac{16}{9} \) into a decimal: \[ \frac{16}{9} \approx 1.777... \] When rounding to the nearest tenth, \( 1.777... \) rounds to \( 1.8 \). Therefore, the sixth term of the sequence, rounded to

To find the sixth term of the sequence defined by ( a_n = \frac{2^n}{n^2} ), substitute ( n = 6 ) into the expression.

Calculating ( a_6 ):

[

a_6 = \frac{2^6}{6^2}

]

First, calculate ( 2^6 ):

[

2^6 = 64

]

Next, compute ( 6^2 ):

[

6^2 = 36

]

Now substitute these values back into the expression for ( a_6 ):

[

a_6 = \frac{64}{36}

]

To simplify ( \frac{64}{36} ), both the numerator and the denominator can be divided by 4:

[

a_6 = \frac{16}{9}

]

Now, we can convert ( \frac{16}{9} ) into a decimal:

[

\frac{16}{9} \approx 1.777...

]

When rounding to the nearest tenth, ( 1.777... ) rounds to ( 1.8 ).

Therefore, the sixth term of the sequence, rounded to