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.
The Correct Answer is B
B.False
Answer:
Given: ABCD is a rectangle.
Prove: The diagonals AC¯¯¯¯¯¯¯¯ and BD¯¯¯¯¯¯¯¯ are congruent.
Match each numbered statement to the correct reason to complete the proof.
PS : i will mark brainliest if they answer the question fully..
Step-by-step explanation:
Given: ABCD is a rectangle.
Prove: The diagonals AC¯¯¯¯¯¯¯¯ and BD¯¯¯¯¯¯¯¯ are congruent.
Match each numbered statement to the correct reason to complete the proof.
PS : i will mark brainliest if they answer the question fully..
Are there any other things that cost money that are in the equation?
