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:
bro can u see the answers?
the answer is -65y+11x/10
Step-by-step explanation:
The given sequence is
a₁, a₂, ...,

Because the given sequence is an arithmetic progression (AP), the equation satisfied is

where
d = the common difference.
The common difference may be determined as
d = a₂ - a₁
The common difference is the difference between successive terms, therefore
d = a₃ - a₂ = a₄ - a₃, and so on..
The sum of the first n terms is

Example:
For the arithmetic sequence
1,3,5, ...,
the common difference is d= 3 - 1 = 2.
The n-th term is

For example, the 10-term is
a₁₀ = 1 + (10-1)*2 = 19
Th sum of th first 10 terms is
S₁₀ = (10/2)*(1 + 19) = 100
Answer:
1/2x or x/2
Step-by-step explanation:
You're taking the number you started with and either multiplying it by 1/2 half or dividing it by 2.