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:
x^2 -x -2
Step-by-step explanation:
Use polynomial long division.
0 + x^2 -x -2
___________________________________
(2x+3) | 2x^3 +x^2 -7x -6
-2x^3 -3x^2
_________
-2x^2 -7x
+2x^2 +3x
__________
-4x -6
+2x +6
_______
0
Since the number is 8 you round up so it's 9.0
Answer:
A
Step-by-step explanation:
LOL
