Answer:
[tex]a_n = 2(4) ^{n-1}[/tex]
Step-by-step explanation:
The sequence shown matches that of a geometric sequence of radius 4. To prove it, divide the term [tex]\frac{a_{n + 1}}{a_n}[/tex] and check that [tex]\frac{a_{n + 1}}{a_n}=4[/tex]
Then the formula that represents this sequence is:
[tex]a_n = a_1(r)^{n-1}[/tex]
Where [tex]a_1[/tex] is the first term of the series = 2 and [tex]r[/tex] is the radius of convergence = 4.
Then the equation is:
[tex]a_n = 2(4) ^{n-1}[/tex]
Answer:
nth term of the sequence is [tex]2^{2n-1}[/tex] or [tex]T_{n}=2.(4)^{n-1}[/tex]
Step-by-step explanation:
The given sequence is 2, 8, 32, 128, 512.........
Or we can rewrite the sequence as [tex]2, 2^{3}, 2^{5},2^{7},2^{9} ,...............nth term.[/tex]
Now the new form of sequence confirms that the sequence is an exponential sequence.
Therefore nth term of the sequence will be [tex]T_{n}=2^{2n-1}[/tex] Or [tex]T_{n}=2.(4)^{n-1}[/tex]
