Let the first term A1 = a not = 1 / 4 and an + 1 = 1 / 2An n be an even number or an + 1 / 4 n be an odd number, denote BN = a (2n-1) - 1 / 4, n = 1,2,3 Let A1 = a not = 1 / 4 and an + 1 = 1 / 2An n be even or an + 1 / 4 n be odd, denote BN = a (2n-1) - 1 / 4, n = 1, 3 Find the value of A2 and A3 Judge whether {BN} is equal ratio sequence and prove it

Let the first term A1 = a not = 1 / 4 and an + 1 = 1 / 2An n be an even number or an + 1 / 4 n be an odd number, denote BN = a (2n-1) - 1 / 4, n = 1,2,3 Let A1 = a not = 1 / 4 and an + 1 = 1 / 2An n be even or an + 1 / 4 n be odd, denote BN = a (2n-1) - 1 / 4, n = 1, 3 Find the value of A2 and A3 Judge whether {BN} is equal ratio sequence and prove it

a2=a1+1/4=a+1/4.
a3=(1/2)a2=a/2+1/8.
b1=a1-1/4≠0.
b=a-1/4=(1/2)a-1/4
=(1/2)[a+1/4]-1/4
=(1/2)[a-1/4]
=(1/2)bn,
{BN} is an equal ratio sequence

The known sequence (an) is an arithmetic sequence with the first term A1 = 1 and tolerance d > 0, and 2A2, A10 and 5a5 are equal proportion sequences
The general term formula of (an)

2A2 = 2 (1 + D) A10 = 1 + 9d5a5 = 5 (1 + 4D) they are in equal proportion, so (1 + 9D) ^ 2 = 5 (1 + 4D) × 2 (1 + D) solution d = 82 / 81 or - 2 / 9 (rounding off) so an = a1 + (n-1) × d = 1-82 / 81 + 82 / 81n = 82 / 81n-1 / 81 (n belongs to positive integer) hope to help you, I use a mobile phone, can't receive follow-up, if you have any questions, please

It is known that the sequence {an}, {BN} is an arithmetic sequence with tolerance of 1, the first terms of which are A1 and B1 respectively, and a1 + B1 = 5, A1, B1 ∈ n *, let CN = ABN (n ∈ n *), then the sum of the first 10 terms of the sequence {CN} is equal to______ ．

When A1 and B1 are 1 and 4, C1 = Ab1 = 4, and the sum of the first 10 terms is 4 + 5 + +12 + 13 = 85; when A1, B1 are 2 and 3, C1 = Ab1 = 4, the sum of the first 10 terms is 4 + 5 + +12 + 13 = 85; when A1, B1 are 4 and 1

It is known that the sequence {an}, {BN} is an arithmetic sequence with tolerance of 1, the first terms of which are A1 and B1 respectively, and a1 + B1 = 5, A1, B1 ∈ n *, let CN = ABN (n ∈ n *), then the sum of the first 10 terms of the sequence {CN} is equal to______ ．

When A1 and B1 are 1 and 4, C1 = Ab1 = 4, and the sum of the first 10 terms is 4 + 5 + +12 + 13 = 85; when A1, B1 are 2 and 3, C1 = Ab1 = 4, the sum of the first 10 terms is 4 + 5 + +12 + 13 = 85; when A1, B1 are 4 and 1, C1 = Ab1 = 4, the sum of the first 10 terms is 4 + 5 + +12 + 13 = 85; when A1, B1 are 3 and 2, C1 = Ab1 = 4, the sum of the first 10 terms is 4 + 5 + +12 + 13 = 85; so the sum of the first 10 items of the sequence {CN} is equal to 85, so the answer is 85

Given that the tolerances of {an}, {BN} are 2 and 3 respectively, and BN ∈ n *, if A1 = B1 = 1, find the general term of {ABN}

An = 1 + (n-1) 2 = 2N-1, BN = 1 + (n-1) 3 = 3n-2, so anbn = (2n-1) (3n-2) = 6n-7n + 2