Let the first n terms and Sn = (- 1) ^ n (2n ^ 2 + 4N + 1) - 1 of sequence {an} 1, find the general term formula an of sequence {an} 2, record BN = (- 1) ^ n / an, find the first n terms and TN of the sequence {BN}

Let the first n terms and Sn = (- 1) ^ n (2n ^ 2 + 4N + 1) - 1 of sequence {an} 1, find the general term formula an of sequence {an} 2, record BN = (- 1) ^ n / an, find the first n terms and TN of the sequence {BN}

Sn=(-1)^n(2n^2+4n+1)-1
Sn-1=(-1)^(n-1)[2(n-1)^2+4(n-1)+1]-1
an=Sn-Sn-1=(-1)^n(4n^2+4n)
bn=1/(4n^2+4n)=1/4[1/n-1/(n+1)]
Calculation of TN by superposition method
b1=……
b2=……
b3=……
If I show you the way,

Given that the general term formula of sequence {an} is an = 2n-49, then when SN is the minimum, the number of terms n ()
A. 1B. 23C. 24D. 25

From an = 2n-49, we can get that the sequence {an} is an arithmetic sequence Sn = − 47 + 2n − 492 × n = N2 − 48n = (n-24) 2-242. Combined with the properties of quadratic function, we can get that when n = 24, the sum has a minimum, so we choose: C

Given that the general term formula of sequence {an} is an = 2n-49, then when SN is the minimum, the number of terms n ()
A. 1B. 23C. 24D. 25

From an = 2n-49, we can get that the sequence {an} is an arithmetic sequence Sn = − 47 + 2n − 492 × n = N2 − 48n = (n-24) 2-242. Combined with the properties of quadratic function, we can get that when n = 24, the sum has a minimum, so we choose: C

Let Sn be the sum of N antecedents of sequence {an} and an = 2n-49, then the value of n is ()
A. 12B. 13C. 24D. 25

From an = 2n-49, we can get that the sequence {an} is an arithmetic sequence ∪ A1 = 2-49 = - 47sn = − 47 + 2n − 492 × n = N2 − 48n = (n-24) 2-242. Combined with the properties of quadratic function, we can get that when n = 24, the sum has a minimum, so we choose C