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

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