Help with Recursion?
Find the third, forth, and fifth terms of the sequence defined recursively by a1 = 7, a2 = 3, and an = an-1 -2an-2.
Given a1= 3, a2 = 6 and an+2 = an+1 -an, find the first four terms of the recursive sequence.
Find the coefficient of a^8b^4 in the expansion of (a + b)^12.
Select the recursion formula for 5, 11, 24, 51,…, where a1 = 5.
Any help is appreciated, I don’t necessarily need the answers – just need to know how to work the problems out so I understand the material.
Thanks again!
Favorite Answer
a1 = 7, a2 =3, an = an-1 – 2an-2
So a3 = a2 – 2a1 = 3 – 2(7) = 3 – 14 = -11
a4 = a3 – 2a2 = -11 – 2(3) = -11 – 6 = -17
a5 = a4-2a3 = -17 – 2(-11) = -17 + 22 = 5
The last one, recursion formula for 5, 11,24,51 is pretty easy.
Look at the numbers.
a1 = 5
a2 = 11 = 2*5 + 1
a3 = 24 = 2*11 + 2
a4 = 51 = 2*24 + 3
So, the formula seems to be
an = 2an-1 + (n -1)
Have to think about the coefficient problem.
