If a 4 + a 5 + a 6 = π / 4, then cos s 9

If a 4 + a 5 + a 6 = π / 4, then cos s 9

According to the properties of arithmetic sequence, A4 + A5 + A6 = 3A5 = π / 4, then A5 = π / 12, so a1 + A9 = 2a5 = π / 6
Then S9 = 9 (a1 + A9) / 2 = 3 π / 4, so coss9 = - 2 arithmetic square root / 2

Given the sequence {an} A1 = 1, an + 1 = Sn (n + 2) / N, (n belongs to n *), it is proved that n ∈ N0 makes Sn > 2007 hold
It is proved that there is a natural number N0, for all n > N0, there is Sn > 2007

S1=a1=1
a(n+1)=S(n+1)-Sn=Sn(n+2)/n
S(n+1)=2Sn(n+1)/n
S2=2S1 *2/1
S3=2S2 *3/2
...
Sn=2S(n-1)*n/(n-1)
Multiply the above formulas to get: SN = 2 ^ (n-1) S1 * n = n * 2 ^ (n-1)
So Sn increases monotonically
S9=9*2^8=2304>2007
Therefore, we only need to take any natural number N0 > = 9

Given that the sequence {an} satisfies A1 = 1 / 2, an + 1 = 2An / (an + 1), it is proved that the inequality 0 ＜ an ＜ an + 1 holds for any n belonging to positive integer

Proof: a (n + 1) = 2An / (an + 1) 1 / a (n + 1) = (an + 1) / (2An) = (1 / 2) (1 / an) + 1 / 21 / a (n + 1) - 1 = (1 / 2) (1 / an) - 1 / 2 = (1 / 2) (1 / an - 1) [1 / a (n + 1) - 1] / (1 / an - 1) = 1 / 2, which is the fixed value