In known sequence an, when A1 = 1 and N ≥ 2, the N-1 power of an = a (n-1) + 3 is obtained

In known sequence an, when A1 = 1 and N ≥ 2, the N-1 power of an = a (n-1) + 3 is obtained

When n ≥ 2, there are: a2-a1 = 3a3-a2 = 3 & # 178; a4-a3 = 3 & # 179 The (n-1) power of an-a (n-1) = 3 is all added, and an-a1 = 1 + 3 + 3 & # 178; + 3 & # 179; + The (n-1) power of + 3 = (3 ^ n-1) / 2An = [(3 ^ n-3) / 2] + 1 (n ≥ 2) gives: [is a piecewise function]

It is known that the sequence {an} satisfies A1 = 1, an is equal to the N-1 power of 3 plus an-1

an=3^(n-1)+a(n-1)
List them one by one, then add them up: (superposition method)
an=a1+3^(N-1)+3^(N-2)+…… +3^1

Simplify the evaluation. 4 (the square of 3a, the square of b-ab) - 5 (- the square of AB + the square of 2a), where a = 1 / 2, B = - 1 / 3
4 (3A's Square b-ab's Square) - 5 (- AB's square + 2A's Square b)

Original formula = 12a & # 178; b-4ab & # 178; + 5ab & # 178; - 10A & # 178;
=12a²b+ab²-10a²
=12*1/4*(-1/3)+1/2*1/9-10*1/4
=-31/9

If the domain of F (2 ^ x) is [- 1,1], then the domain of F (log2x) is
Log is the logarithm of base X

-1