In the expansion of (1 + 2x) ^ n, let the sum of binomial coefficients of odd terms be the sum of the first n terms of an sequence and be Sn, then what is Liman / Sn equal to

In the expansion of (1 + 2x) ^ n, let the sum of binomial coefficients of odd terms be the sum of the first n terms of an sequence and be Sn, then what is Liman / Sn equal to

The sum of odd binomial coefficients is 2 ^ (n-1), which is an equal ratio sequence, so Sn = 2 ^ n-1, so Liman / Sn = 1 / 2

Let the first n terms of sequence {an} and Sn = 4 / 3an - {(1 / 3) * 2 ^ n + 1} + 2 / 3
Find the general term of the sequence

By subtracting Sn = 4 / 3an - {(1 / 3) * 2 ^ n + 1} + 2 / 3, s (n-1) = 4 / 3A (n-1) - {(1 / 3) * 2 ^ (n-1) + 1} + 2 / 3, we get SN-S (n-1) = 4 / 3 (an-a (n-1)) - (1 / 3) * 2 ^ n + (1 / 3) * 2 ^ (n-1), that is, an = 4 / 3 (an-a (n-1)) - (1 / 3) * 2 ^ (n-1), an - (1 / 2) * 2 ^ n = 4A (n-1), which is rewritten as an + (1 / 2) * 2

Given the first n terms of sequence {an} and Sn = (n ^ 2 + n) * 3 ^ n, (1) find LIM (n - > OO) (an / Sn)

An / Sn = [SN-S (n-1)] / Sn = 1 - [S (n-1)] / Sn = 1 - {[(n-1) ^ 2 + (n-1)] * 3 ^ (n-1)} / [(n ^ 2 + n) * 3 ^ n] = 1 - {[(n-1) ^ 2 + (n-1)]} / [(n ^ 2 + n) * 3] in the following fraction of the formula, the number of the highest order terms of the numerator and denominator is 2, so its limit is equal to the ratio of the coefficients of the highest order terms

LIM (n →∞) [(n ^ 3-1) / (3 * n ^ 2 + n) - an-b] = 0 find the value of a and B

lim(n→∞) [(n^3-1）/(3*n^2+n)-an-b]
=lim(n→∞) [(1-3a)n^3-(a+3b)n^2-bn-1]/(3*n^2+n)]=0
therefore
1-3a = 0, and a + 3B = 0
thus
a=1/3,b=-1/9.