Question

Please help very urgent

Answer 1

Answer:

(a)

$T_n = 3(-4)^{n-1}$

(b)

$\sum\limits^n_1 { 3(-4)^{n-1}$

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:

$r = \frac{-12}{3} = \frac{48}{-12} = \frac{-192}{48} = \frac{768}{-192}$

$r = -4$

The explicit formula is calculated using the n term of an GP.

$T_n = ar^{n-1}$

$T_n = 3 * (-4)^{n-1}$

$T_n = 3(-4)^{n-1}$

Solving (b): Summation notation.

This implies that the sum of the sequence.

To do this, we write Tn inside the summation sign

i.e.

$\sum\limits^n_1 {T_n}$

Substitute values for Tn

$\sum\limits^n_1 { 3(-4)^{n-1}$