The given sequence is an arithmetic series with a common difference of 4, divided by n^2.

The sum of an arithmetic series can be found using the formula:
S_n = n/2 * (a_1 + a_n), where a_1 is the first term and a_n is the last term.

In this case, a_1 = 6 and a_n = 4n - 2.

So, S_n = n/2 * (6 + 4n - 2) = 2n^2 + 2n - n = 2n^2 + n.

The nth term of the sequence a_n is then S_n / n^2 = (2n^2 + n) / n^2 = 2 + 1/n.

The limit of the sequence as n approaches infinity is then the limit of 2 + 1/n as n approaches infinity, which is 2.

So, the limit of the sequence is 2.

Question

The given sequence is an arithmetic series with a common difference of 4, divided by n^2.

The sum of an arithmetic series can be found using the formula: 
S_n = n/2 * (a_1 + a_n), where a_1 is the first term and a_n is the last term.

In this case, a_1 = 6 and a_n = 4n - 2.

So, S_n = n/2 * (6 + 4n - 2) = 2n^2 + 2n - n = 2n^2 + n.

The nth term of the sequence a_n is then S_n / n^2 = (2n^2 + n) / n^2 = 2 + 1/n.

The limit of the sequence as n approaches infinity is then the limit of 2 + 1/n as n approaches infinity, which is 2.

So, the limit of the sequence is 2.

Knowee AI · Accepted Answer

The given sequence is an arithmetic series with a common difference of 4, divided by n^2.

The sum of an arithmetic series can be found using the formula: 
S_n = n/2 * (a_1 + a_n), where a_1 is the first term and a_n is the last term.

In this case, a_1 = 6 and a_n = 4n - 2.

So, S_n = n/2 * (6 + 4n - 2) = 2n^2 + 2n - n = 2n^2 + n.

The nth term of the sequence a_n is then S_n / n^2 = (2n^2 + n) / n^2 = 2 + 1/n.

The limit of the sequence as n approaches infinity is then the limit of 2 + 1/n as n approaches infinity, which is 2.

So, the limit of the sequence is 2.

Find the limit of the sequence given by 𝑎𝑛=6+10+14+...+(4𝑛−2)𝑛2a n = n 2 6+10+14+...+(4n−2) , (where 𝑛∈𝑁∖{0}n∈N∖{0}).