Compute the first 5 terms of the sequence: {b(n)}n=1�¨�‡ = n�ã6n + (5/6)^n.
((that is:(Bsub(n))from n=1 to infinity = nth root of 6n + (5/6)raised to n.))
If the sequence converges, find the limit.
To compute the first 5 terms of the sequence, we substitute the values of n from 1 to 5 into the given formula.
1. For n = 1:
b(1) = 1 * 6 + (5/6)^1 = 6 + 5/6 ≈ 6.833
2. For n = 2:
b(2) = 2 * 6 + (5/6)^2 = 12 + 25/36 ≈ 12.694
3. For n = 3:
b(3) = 3 * 6 + (5/6)^3 = 18 + 125/216 ≈ 18.579
4. For n = 4:
b(4) = 4 * 6 + (5/6)^4 = 24 + 625/1296 ≈ 24.481
5. For n = 5:
b(5) = 5 * 6 + (5/6)^5 = 30 + 3125/7776 ≈ 30.401
Now, let's check if the sequence converges. For a sequence to converge, the terms must approach a specific value as n goes to infinity.
To find the limit, we can observe the behavior of the terms as n increases.
If we take the limit of the sequence as n approaches infinity, it simplifies to:
lim(n→∞) b(n) = lim(n→∞) (6n + (5/6)^n)
The first term in the expression, 6n, clearly goes to infinity as n increases.
The second term, (5/6)^n, approaches zero as n goes to infinity because any number less than 1 raised to a large power tends to zero.
Therefore, the overall limit is infinity.
In summary, the first 5 terms of the sequence {b(n)} are approximately 6.833, 12.694, 18.579, 24.481, and 30.401. The sequence diverges and the limit is infinity.