Question

Find the sum of a finite geometric sequence from n = 1 to n = 5, using the expression −3(4)^(n − 1).

Answer 1

@ganeshie8

Answer 2

@aum

Answer 3

@gswag98

Answer 4

is this like distributing

Answer 5

I am not sure, but I don't think so.

Answer 6

@camerondoherty

Answer 7

They give you an equation:
f(x)=-3(4)^(n-1)
So what you need to do is sub n for 1, 2, 3, 4, 5 and solve for each. The you will get the numbers in your sequence...
Here's an example:
For n=1
f(x)=-3(4)^(n-1)
f(1)=-3(4)^(1-1)
f(1)=-3(4)^0
f(1)=-12^0
f(1)=-1

Answer 8

So you do that with all the different vales of n that they tell you to solve for

Answer 9

but dont put f(x) put g(x) to show it is a geometric sequence

Answer 10

Wow, smart and beautiful, thank you!

Answer 11

Lol thanx c;

Answer 12

@aum can u help me out on my question when you get the time

Answer 13

−3 * (4)^(n − 1) ----- (1)
Put n = 1, -3 * 4^0 = -3 * 1 = -3
To find the other numbers, you can put n = 2, 3, 4 and 5 into (1) and do the calculation. But an easier method would be to notice that every time n increases by 1 it is equivalent to multiplying by 4 because the base is 4 and n occurs in the exponent.
So the second number is: first number * 4 = -3 * 4 = -12
Third number is: -12 * 4 = -48
Fourth number is: -48 * 4 = -192
Fifth number is: -192 * 4 = -768

Add: -3 + (-12) + (-48) + (-192) + (-768) =
-(3 + 12 + 48 + 192 + 768) = ?