Sequence {an} = 3A (n-1) + 5, A1 = 2, find an

Sequence {an} = 3A (n-1) + 5, A1 = 2, find an

An = 3A (n-1) + 5 (n ＞ 1) an = 3A (n-1) + (15 / 2) - (5 / 2) an + (5 / 2) = 3A (n-1) + (15 / 2) an + (5 / 2) = 3 [a (n-1) + (5 / 2)] then an + (5 / 2) is an equal ratio sequence, the first term is 2 + (5 / 2) = 4.5, and the common ratio is 3an + (5 / 2) = 4.5 × 3 ^ (n-1) an = 4.5 × 3 ^ (n-1) - 2.5

Finding an with known sequence an A1 = 1 and 3a (n + 1) - 3A n = 5

This is an arithmetic sequence with a first term of 1 and a tolerance of 5 / 3
an=1+(n-1)*5/3=(5n-2)/3

The known sequence an satisfies: A1 = 1-3k, an = 4 ^ n-1-3a (n-1)
(1) Judge whether the sequence an-4 ^ n / 7 is an equal ratio sequence;
(2) If the sequence an is an increasing sequence, the value range of K is obtained;

An = 4 ^ n-1-3a (n-1) an-4 ^ n / 7 = 3 [a (n-1) - (4 ^ n-1) / 7] [an-4 ^ n / 7] / [a (n-1) - (4 ^ n-1) / 7] = 3 into an equal ratio sequence
An = A1 * 3 ^ (n-1) = (1-3k) * 3 ^ (n-1) when A1 is less than 0, the value range of K is greater than 1 / 3