From the given recurrence, it follows that

and so on down to the first term,

(Notice how the exponent on the 2 and the subscript of <em>a</em> in the first term add up to <em>n</em> + 1.)
Denote the remaining sum by <em>S</em> ; then

Multiply both sides by 2 :

Subtract 2<em>S</em> from <em>S</em> to get

So, we end up with

Answer:
5
Step-by-step explanation:
Answer:
p = 9/4 or 2.25
Step-by-step explanation:
4p + 1 = 10
-1 -1
4p = 9
Divide 4 on both sides
p = 9/4 or 2.25
Answer:
is a sas theorem.
Step-by-step explanation:
Answer:
x=80
Step-by-step explanation:
im going to assume this is perimeter.
50+50=100
260-100=160
160÷2=80