Question

consider the sequence :
a1= sqrt 12
a2= sqrt(12+sqrt12)
a3= sqrt(12+sqrt(12+sqrt12))
a4= sqrt(12+sqrt(12+sqrt(12+sqrt12)))
.
.
.
a) find a recursion function for a(n+1) (term)
b) Assuming that the sequence converges, find the limit.

Answer 1

well, from the pattern, it appears that a next term is equal to the sqrt of the now term+12

Answer 2

\[A_{n+1}=\sqrt{A_n+12}\]

right?

Answer 3

yes

Answer 4

rewriting the sqrt as an exponent might be helpful ... but might not :)

\[A_{n+1}=(A_n+12)^{1/2}\]

Answer 5

should we try to develop an explicit formula for An?

Answer 6

no, i found the limit, it's 4. Thx a lot. :)

Answer 7

whew!! thats good :)