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It is known that in the arithmetic sequence an, A4 = 4, the first 10 terms and S10 = 10, BN = (1 / C) are to the power of an, (C is a positive constant.) (1) Find an (2) It is proved that BN is an equal ratio sequence (3) Finding the first n terms and TN of sequence BN
（1）S10=(a4+a7)*5,a4+a7=2
a7=-2,d=-2,an=12-2n
(2) bn=(1/c)^(12-2n)
b(n+1)/bn=c^2
BN is equal ratio
（3）Tn=(1/c)^10+(1/c)^8+…… +(1/c)^(12-2n)
=((1/c)^10*(1-c^2n))/(1-c^2)
Given that the sum of the first n terms of the sequence {an} is Sn, and an = n power of N2, then SN=
Sn=1*2+2*2^2+3*2^3+.+n*2^n ,
Multiply both sides by 2 to get 2Sn = 1 * 2 ^ 2 + 2 * 2 ^ 3 + 3 * 2 ^ 4 +. + (n-1) * 2 ^ n + n * 2 ^ (n + 1),
If you subtract the two expressions, you get
Sn=2Sn-Sn=-1*2-2^2-2^3-.-2^n+n*2^(n+1)
= -[2^(n+1)-2]+n*2^(n+1)
=2+(n-1)*2^(n+1) .
Let the common ratio of the equal ratio sequence {an} be q, and the sum of the first n terms be SN. If Sn + 1, Sn, Sn + 2 become the equal difference sequence, then the value of Q is () I know the answer is - 2, but I calculate - 2 and 1. Why can't 1?
When the common ratio q = 1, Sn = n, Sn + 1 = n + 1, Sn + 2 = n + 22sn = 2n, Sn + 1 + Sn + 2 = 2n, Sn + 1 = 2n + 32sn ≠ Sn + 1 + Sn + 2, it does not satisfy the meaning of the problem, so the common ratio Q ≠ 1sn + 1, Sn, Sn + 2 become the arithmetic sequence, then 2Sn = Sn + 1 + Sn + 22a1 (Q ^ n-1) / (Q-1) = A1 [Q ^ (n + 1) - 1] / (Q-1) + A1 [...]
If s (n + 1), Sn and S (n + 2) are equal difference sequences, then the value of Q?
Q = 1 or q = - 2
Sn=(a1-a1*q^n)/(1-q)
Sn+1=(a1-a1*q^n+1)/(1-q)
Sn+2=(a1-a1*q^n+2)/(1-q)
(a1-a1*q^n)*2=a1-a1*q^(n+1)+a1-a1*q^(n+2)
2=q+q^2
Q = 1 or q = - 2
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