Question

What is the `6^(th)` term in the recursive formula sequence: `a_1=1`, `a_n=(a_(n−1))^2−10`?

Answer 1

`a_6=640644240524391`

Explanation:

`a_1=1, a_n=(a_(n-1))^2-10`

When `n=2, a_2=(a_1)^2-10=1^2-10=-9`

When `n=3, a_3=(a_2)^2-10=(-9)^2-10=71`

When `n=4, a_4=(a_3)^2-10=71^2-10=5031`

When `n=5, a_5=(a_4)^2-10=5031^2-10=25310951`

When `n=6, a_6=(a_5)^2-10=640644240524391`

Answer 2

the `6^(th)` term is `640644240524391`

Explanation:

We have a recursive sequence defined by:

`{ (a_1=1), (a_n=(a_(n−1))^2−10) :}`

Then we set `n` accordingly and compute further terms:

# {: (n=1, => a_1=1, ,"(by definition)"), (n=2, => a_2=(a_1)^2−10, = 1-10, = -9), (n=3, => a_3=(a_2)^2−10, = 81-10, = 71), (n=4, => a_4=(a_3)^2−10, = 5041-10, = 5031), (n=5, => a_5=(a_4)^2−10, = 25310961-10, = 25310951), (n=6, => a_6=(a_5)^2−10, = 640644240524401-10, = 640644240524391) :} #

Thus, the `6^(th)` term is `640644240524391`