Urgently need the general formula of the sequence an
help! The general formula of the sequence an
S(n 1) = 4an 2..........(A)
Sn = 4a(n-1) 2.......(B)
(A)-(B) Obtain, a(n 1) = 4an - 4a(n-1)
Transfer the terms, a(n 1) - 2an = 2an - 4a(n-1) = 2[an - a(n-1)]
Suppose bn = a(n 1) - 2an
Then, bn = 2b(n-1) q = 2
According to the question, S2 = a1 a2 = 4a1 2
Because a1 = 1 a2 = 5
So, b1 = a2 - 2a1 = 3
So, bn = b1*q^(n-1) = 3 * 2^(n-1)
That is a(n 1) - 2an = 3 * 2^(n-1)
2[an - 2a(n-1)] = 3 * 2^(n-2) * 2 = 3 * 2^(n-1)
2^2*[a(n-1) - 2a(n-2)] = 3 * 2^(n-3) * 2^2 = 3 * 2^(n-1)
: :
: :
: :
2^(n-1)*(a2 - 2a1) = 3 * 2^(n-1)
The above are n formulas, add the above formulas,
Get a(n 1) - 2^n*a1 = 3 * 2^(n-1) * n
a(n 1) - 2^n = 3 * 2^(n-1) * n
a(n 1) = 2^n 3 * 2^(n-1) * n
So an = 2^(n-1) 3 * 2^(n-2) * (n-1)
= 2^(n-2) * (3n-1)
The formula of the general term of the sequence
Is your title wrong? If a2 a4=5/4, a1 a5=1/4, a1 and q have no solution. .
a1*a5=1/4 is almost the same.
In that case, there are a2*a4=a1*a5=1/4, a2 a4=5/4
Solving the system of equations, we get: a2=1, a4=1/4 or a2=1/4, a4=1 (discarded because 0 = 4-(1/2)^(n-2)-n*(1/2)^*(n-1)
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