Question 1022414
You can use a_{n} to represent *[tex \large a_n], a_{n-1} to represent *[tex \large a_{n-1}].


For the first one, what it is saying is that *[tex \large a_n] (the nth term of the sequence) is the square of the previous term *[tex \large a_{n-1}], plus 1:

*[tex \large a_1 = 1]
*[tex \large a_2 = a_1^2 + 1 = (1)^2 + 1 = 2]
*[tex \large a_3 = a_2^2 + 1 = (2)^2 + 1 = 5]
*[tex \large a_4 = a_3^2 + 1 = (5)^2 + 1 = 26]
*[tex \large a_5 = a_4^2 + 1 = (26)^2 + 1 = 677]


The second one says that *[tex \large a_n] is equal to the quantity *[tex \large n^2 - 2n], for all integers *[tex \large n \ge 1].


*[tex \large a_1 = 1^2 - 2 \cdot 1 = -1]
*[tex \large a_2 = 2^2 - 2 \cdot 2 = 0]
*[tex \large a_3 = 3^2 - 2 \cdot 3 = 3]
*[tex \large a_4 = 4^2 - 2 \cdot 4 = 8]
*[tex \large a_5 = 5^2 - 2 \cdot 5 = 15]