Given the geometric sequence that begins with the terms 2,1,... What is the sum of the first 12 terms?

The general term of a geometric series is arn−1, where r is the ratio, and a is the first term. In your case, a=2–√ and r=3–√. The 12th term is then

2–√(3–√)11=2436–√

The sum of the first 12 terms is

a∑k=011rk=ar12−1r−1=2–√7283–√−1=364(2–√+6–√)

Note that I used 35=243 and 36=729 in the above.

EDIT

Note that

2–√7283–√−1=2–√7283–√−13–√+13–√+1=2–√(3–√+1)728(3–√)2−12=(6–√+2–√)7283−1

Step-by-step explanation:

b. 10=9.1+0.2x

step-by-step explanation:

that's it

there are many solutions.

step-by-step explanation:

step 1: simplify both sides of the equation.

4x+6=4x^2+10x−4

step 2: subtract 4x^2+10x-4 from both sides.

4x+6−(4x^2+10x−4)=4x^2+10x−4−(4x^2+10x−4)

−4x^2−6x+10=0

step 3: factor left side of equation.

−2(2x+5)(x−1)=0

step 4: set factors equal to 0.

x=-5/2 or x=1