1003107309 | math4u.vsb.cz

SubArea:

Introduction to sequences

Level:

Project ID:

1003107309

Source Problem:

Accepted:

1

Clonable:

1

Easy:

0

We are given the sequence

{(\log 2^{n})}_{n = 1}^{\infty}

. Find the recursive formula of such sequence.

a_{1} = \log 2; a_{n + 1} = a_{n} + \log 2, n \in N

a_{1} = \log 2; a_{n + 1} = a_{n} \cdot \log 2, n \in N

a_{1} = \log 2; a_{n + 1} = a_{n} - \log 2, n \in N

a_{1} = \log 2; a_{n + 1} = a_{n} + \log 2^{n}, n \in N