Given that an = 32n − 11 (n ∈ n *), the sum of the first n terms of the sequence {an} is Sn, then the minimum value of n with Sn > 0 is () A. 10B. 11C. 12D. 13

Given that an = 32n − 11 (n ∈ n *), the sum of the first n terms of the sequence {an} is Sn, then the minimum value of n with Sn > 0 is () A. 10B. 11C. 12D. 13


From an = 32n − 11 (n ∈ n *), a1 + A10 = A2 + A9 = =A5 + A6 = 0, a11 > 0, S9 < 0, S10 = 0, S11 > 0, so that the minimum value of n with Sn > 0 is 11



Given the sequence {an}, if 1 / A1A2 + 1 / a2a3 + +1 / anan-1 = n / Anan + 1, prove {an} is arithmetic sequence


On the basis of the original formula, write another equation with the same structure until an + 2. Subtract the original formula to get: 1 / (an + 1) an = n + 1 / (an + 1) (an + 2) - N / Anan + 1 We assume that the tolerance is d. then the left side of the original formula = 1 / D (1 / A1-1 / A2...) =1/d(1/a1-1/...



Given that the sequence {an} satisfies A1 = 1, an + 1 = an / 1 + 2An (n belongs to n *), if A1A2 + a2a3 + +Anan + 1 > 16 / 33
The known sequence {an} satisfies A1 = 1, an + 1 = an / 1 + aan (n belongs to n *)
Ask if A1A2 + a2a3 + +Ana (n + 1) > 16 / 33, find the value range of n


An + 1 = an / 1 + 2An
1/an+1-1/an=2
1/an=1+(n+1)*2=2n+3
an=1/[2n+3]
a1a2+a2a3+…… +anan+1=1/2[1/a1-1/a2+1/a2-a/a3+…… -1/(2n+3)]=1/2-1/2(2n+3)>16/33
n>15