If an = 2n (n ∈ n *), then A1A2 + a2a3 + a3a4 + +anan+1=

If an = 2n (n ∈ n *), then A1A2 + a2a3 + a3a4 + +anan+1=

Refer to Baidu,] an = 2n, that is 246 8 10 12 14 16 A1A2 + +Anan + 1 = an, i.e. 8 24 48 80 120 168 So, Sn = 4 × n (n + 1) (2n + 1) / 6 + 4 × n (n + 1) / 2 = 2n (n + 1) (2n + 1) / 3 + 2n (n + 1) = 2n (n + 1) (2n

The first term of sequence {an} is 2, and for any n ∈ n *, there is 1 / A1A2 + 1 / a2a3 +... + 1 / Anan + 1 = n / a1an + 1, and the sum of the first 10 terms of sequence {an} is 110
(1) Verification: the sequence {an} is an arithmetic sequence;
(2) Let CN = an &; (1 / 2) ^ n, find the first n terms and TN of sequence {CN};
(3) If there is n ∈ n *, such that an ≤ (n + 1) λ holds, the minimum value of real number λ is obtained

According to the meaning of the title, 1 / A1A2 + 1 / a2a3 1 / anan-1 = (n-1) / a1an (1) the original formula - 1 obtains 1 / Anan + 1 = n / a1an + 1 - (n-1) a1an, arranges 2 = Nan - (n-1) an + 1, divides both sides of n (n-1) at the same time, obtains 2 / n (n-1) = an / (n-1) - an + 1 / N2 / (n-1) - 2 / N = an / (n-1) - an + 1 / N (an + 1 - 2) / N = (an - 2) / (n-1) = ...

If the sequence {an} satisfies A1 = 1,1 / 2An = 1 / 2An + 1 (n ∈ n ＊), if A1A2 + a2a3 +... + Anan + 1 > 16 / 33, the value of n is obtained

If 1 / 2An = 1 / 2A (n-1) + 1, multiply both sides by 2 to get 1 / an = 1 / a (n-1) + 2, then (1 / an) can be regarded as the arithmetic sequence A1 = 1, and 1 / an = 2N-1 an = 1 / (2n-1) A1 * A2 + A2 * A3 +... + an * an + 1 = 1 / 1 * 3 + 1 / 3 * 5 +. 1 / (2n-1) * (2n + 1) > 16 / 33 1 / 1 * 3 + 1 / 3 * 5 +. 1 / (2n-1) * (2n +...)