In triangle ABC, a = 2, B = 1, find the value range of angle B

In triangle ABC, a = 2, B = 1, find the value range of angle B

(0°,30°]
If CoSb = [A & # 178; + C & # 178; - B & # 178;] / (2Ac) = [3 + C & # 178;] / (4C) = 3 / (4C) + (C / 4) ≥ √ 3 / 2, then B ∈ (0, π / 6]

The first few terms of arithmetic sequence and sntn, Sn / TN = 2n / 3N + 1 for an / BN

The first few terms of arithmetic sequence and sntn, Sn / TN = 2n / 3N + 1
Then an / BN = [A1 + a (2n-1)] / 2 △ [B1 + B (2n-1)] / 2 = {[A1 + a (2n-1)] / 2} × (2n-1) / {[B1 + B (2n-1)] / 2} × (2n-1)
=S(2n-1)÷T(2n-1)=2(2n-1)÷[3(2n-1)+1]=(4n-2)/(6n-2)=(2n-1)/(3n-1)

If the sum of the first n terms of {an} and {BN} is Sn and TN respectively, and Sn / TN = (2n + 1) / (3N + 2), then: A6 / B6 =?

The sum of the first n terms of the arithmetic sequence {an} and {BN} is Sn and TN respectively, and Sn / TN = (2n + 1) / (3N + 2), then: A6 / B6 =? You should pay attention to; what's the relationship between A6 and Sn; the following is the arithmetic sequence of comparison skills: S11 = a1 + A2 +. + A5 + A6 + A7 +. + A10 + a11 = (a6-5d) + (a6-4d) +. + (a6-D) + (A6 +) +

Sum: 11 × 4 + 14 × 7 + +1(3n-2)×(3n+1)= ___ ．

Let Sn = 11 × 4 + 14 × 7 + +1 (3n-2) × (3N + 1), then 3Sn = 31 × 4 + 34 × 7 + +3(3n-2)×(3n+1)=1-14+14-17+… +1 (3n-2) - 1 (3N + 1) = 1-1 (3N + 1) = 3N (3N + 1), so Sn = n (3N + 1)