The sum of the first n terms of the arithmetic sequence an is Sn, an = 5, S5 = 35, and the sequence BN satisfies an = log2bn Hope the master gives a detailed answer, I am very poor in mathematics, O (∩)_ Thank you It's A2 = 5. Sorry, I sent it wrong

The sum of the first n terms of the arithmetic sequence an is Sn, an = 5, S5 = 35, and the sequence BN satisfies an = log2bn Hope the master gives a detailed answer, I am very poor in mathematics, O (∩)_ Thank you It's A2 = 5. Sorry, I sent it wrong

Because the sequence {an} is an arithmetic sequence, A3 = (a1 + A5) / 2
Because S5 = 5 * [(a1 + A5) / 2] = 35, that is, S5 = 5A3 = 35, so A3 = 7
Because A2 = 5, the tolerance d = 2
So an = A2 + (n-2) * d = 5 + 2 (n-2) = 2n + 1
Because an = log2 (BN), BN = 2 ^ (an) = 2 ^ (2n + 1)
So TN = 2 ^ 3 + 2 ^ 5 + 2 ^ 7 + +2^(2n+1)=8*(1-4^n)/(1-4)=(8/3)*(4^n-1).
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Let the probability distribution law of random variables (x, y) be as shown in the figure, and find: (1) the marginal distribution law of X; (2) the distribution law of Z = x + y
Y\X -1 0 10 0 0.2 0.21 0.2 0.3 0.1

(1) Edge distribution law of X
X -1 0 1
P 0.2 0.5 0.3
(2) The distribution law of Z = x + y
Z -1 0 1 2
P 0 0.4 0.5 0.1
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If you are satisfied with the answer,