Answer:
(a)
[tex]T_n = 3(-4)^{n-1}[/tex]
(b)
[tex]\sum\limits^n_1 { 3(-4)^{n-1}[/tex]
Step-by-step explanation:
Given
The above sequence
Solving (a): The explicit formula
The given sequence is a geometric sequence because the common ratio (r) is:
[tex]r = \frac{-12}{3} = \frac{48}{-12} = \frac{-192}{48} = \frac{768}{-192}[/tex]
[tex]r = -4[/tex]
The explicit formula is calculated using the n term of an GP.
[tex]T_n = ar^{n-1}[/tex]
[tex]T_n = 3 * (-4)^{n-1}[/tex]
[tex]T_n = 3(-4)^{n-1}[/tex]
Solving (b): Summation notation.
This implies that the sum of the sequence.
To do this, we write Tn inside the summation sign
i.e.
[tex]\sum\limits^n_1 {T_n}[/tex]
Substitute values for Tn
[tex]\sum\limits^n_1 { 3(-4)^{n-1}[/tex]
