Question

Can somebody help me find an expression for the series 2, 2.5, 2.75, 2.875, ... ?

Answer 1

+0, +1/2, +3/4, + 5/8 maybe? id have to dbl chk that

Answer 2

nah, that 2.625 in the end .... bummer

Answer 3

7/8 on the end

Answer 4

looks like the underside is going by powers of 2 tho

Answer 5

the top, 0,1,3,7 ; not alot to work off of

Answer 6

1/2 = .5

1/2 + 1/4 = .75

1/2 + 1/4 + 1/8 = .875

Answer 7

An = A{n-1} + 1/2^n

Answer 8

\[\left \{2+\frac{2^n-1}{2^n}:n\in\mathbb{N}\right \}\]

Answer 9

do we start at n=0 or n=1?

Answer 10

A1 = 1 + 1/2^(1-1) = 2
A2 = 1 + 1/2^(1-1) + 1/2^(2-1) = 2.5
A3 = 1 + 1/2^(1-1) + 1/2^(2-1) + 1/2^(3-1)= 2.75

Answer 11

Adding powers of 5 with decimal points of powers of 10

Answer 12

An = 1 + 2^0 + 2^-1 + 2^-2 + .... +2^-(n-1)

Answer 13

that works if we can get the closed form from it which should be 1 + geometric sum

Answer 14

(1-(1/2)^n)/(1-(1/2)) right?

Answer 15

that works

Answer 16

\[A_n=1+\frac{1-\frac{1}{2}^n}{1-\frac{1}{2}}\]
\[A_n=1+\frac{1-\frac{1}{2}^n}{\frac{1}{2}}\]
\[A_n=1+2(1-\frac{1}{2}^n)\]
\[A_n=1+2-2(\frac{1}{2})^n)\]
\[A_n=3-2(\frac{1}{2})^n)\]

Answer 17

hmmm.... its good up to:

\[A_n=1+2(1-\frac{1}{2^n})\]

Answer 18

im assuming we start at A1

Answer 19

\[A_n=1+2-\frac{2}{2^n}\]
\[A_n=3-2^{1-n}\]

that one works for me