the limit from n->infinity of the summation of [2+(3/n)k]^2(3/n) when k=1

Question

the limit from n->infinity of the summation of [2+(3/n)k]^2(3/n) when k=1

Answer 1

To find the limit of the given summation, we need to apply the limit as n approaches infinity to the given expression.

Let's break down the given expression: [2 + (3/n)k]^2 * (3/n)

Firstly, let's consider the expression inside the square brackets: [2 + (3/n)k].

As k is a constant and n approaches infinity, the term (3/n)k tends to 0. Therefore, [2 + (3/n)k] approaches 2.

Now, let's simplify the overall expression: [2 + (3/n)k]^2 * (3/n)

Substituting [2 + (3/n)k] with 2, we have:
2^2 * (3/n) = 4 * (3/n) = 12/n

Now, let's calculate the limit of this expression as n tends to infinity:

lim(n→∞) 12/n = 0

Therefore, the limit of the given summation as n approaches infinity is 0.