Not an expertise on infinite sums but the most straightforward explanation is that infinity isn't a number.
Let's see if there are anything we missed:
∞
Σ 2^n=1+2+4+8+16+...
n=0
We multiply (2-1) on both sides:
∞
(2-1) Σ 2^n=(2-1)1+2+4+8+16+...
n=0
And we expand;
∞
Σ 2^n=(2+4+8+16+32+...)-(1+2+4+8+16+...)
n=0
But now, imagine that the expression 1+2+4+8+16+... have the last term of 2^n, where n is infinity, then the expression of 2+4+8+16+32+... must have the last term of 2(2^n), then if we cancel out the term, we are still missing one more term to write:
∞
Σ 2^n=-1+2(2^n)
n=0
If n is infinity, then 2^n must also be infinity. So technically, this goes back to infinity.
Although we set a finite term for both expressions, the further we list the terms, they will sooner or later approach infinity.
Yep, this shows how weird the infinity sign is.
Answer:
4 sec
Step-by-step explanation:
let y = 0 which is the height when the rocket hits the ground
-3t^2 + 6t + 24 = 0
-3(t^2 - 2t - 8) = 0
-3(t - 4)(t + 2) = 0
t = 4 or t = -2
time cannot be negative, so t = 4 sec
Answer: She is incorrect because 0 + 0 = <u>0</u>
Answer:
b-3x+19
Step-by-step explanation:
you have to open the bracket so 3*x=3x
3*3=9
therefore;3x+9+10=3x+19
