Question

Evaluating each geometric series described. Find the sum A1=-3, An=-196608, r=4 Answer choices: A) 1 B)-285886 C)-262143 D)-339855 S=

Answer 1

Answer:

(C)-262143

Step-by-step explanation:

For a geometric series, the nth term

$A_n=ar^{n-1}$ where a= first term, r=common ratio.

If $A_1=a=-3$ , r=4 and $A_n=-196608$

then from $A_n=ar^{n-1}$

-196608=-3 X $4^{n-1}$

$4^{n-1}=\frac{-196608}{-3} =65536\\4^{n-1}=4^8$

Since the bases are the same, the powers are equal

n-1=8

n=8+1=9

Therefore the Sum of the geometric series

$S_n=\frac{a(r^n-1)}{r-1}$ (This form is used because r>1)

$S_n=\dfrac{-3(4^{9}-1)}{4-1}$

$S_n=\dfrac{-3(262144-1)}{3}=\dfrac{-3(262143)}{3}=-262143$