Evaluate The Limit: Lim ⁡ N → ∞ 3 N ∑ K = 1 N ( 2 + 3 K N ) 4 \lim_{n \rightarrow \infty} \frac{3}{n} \sum_{k=1}^n\left(2+\frac{3 K}{n}\right)^4 Lim N → ∞ ​ N 3 ​ ∑ K = 1 N ​ ( 2 + N 3 K ​ ) 4

Introduction

In this article, we will delve into the world of mathematical limits and explore the evaluation of a specific summation. The limit in question is $\lim_{n \rightarrow \infty} \frac{3}{n} \sum_{k=1}^n\left(2+\frac{3 k}{n}\right)^4$. This limit involves a summation of a polynomial expression, and our goal is to simplify and evaluate it as $n$ approaches infinity.

Understanding the Summation

The given summation is $\sum_{k=1}^n\left(2+\frac{3 k}{n}\right)^4$. This is a sum of a polynomial expression, where each term is raised to the power of 4. The expression inside the summation can be rewritten as $\left(\frac{n}{n} + \frac{3k}{n}\right)^4 = (1 + \frac{3k}{n})^4$. This form will be useful in our evaluation of the limit.

Evaluating the Limit

To evaluate the limit, we can start by rewriting the expression as $\frac{3}{n} \sum_{k=1}^n (1 + \frac{3k}{n})^4$. This can be further simplified by expanding the polynomial expression using the binomial theorem.

Binomial Expansion

Using the binomial theorem, we can expand the polynomial expression as follows:

$(1 + \frac{3k}{n})^4 = 1 + \frac{12k}{n} + \frac{54k^2}{n^2} + \frac{108k^3}{n^3} + \frac{81k^4}{n^4}$

Simplifying the Summation

Now, we can substitute this expansion back into the original summation:

$\sum_{k=1}^n (1 + \frac{12k}{n} + \frac{54k^2}{n^2} + \frac{108k^3}{n^3} + \frac{81k^4}{n^4})$

Evaluating the Summation

To evaluate the summation, we can use the formula for the sum of a geometric series. The sum of a geometric series is given by:

$\sum_{k=1}^n ar^k = a \frac{1-r^{n+1}}{1-r}$

Applying the Formula

We can apply this formula to each term in the summation:

$\sum_{k=1}^n \frac{12k}{n} = \frac{12}{n} \sum_{k=1}^n k = \frac{12}{n} \frac{n(n+1)}{2} = 6(n+1)$

$\sum_{k=1}^n \frac{54k^2}{n^2} = \frac{54}{n^2} \sum_{k=1}^n k^2 = \frac{54}{n^2} \frac{n(n+1)(2n+1)}{6} = 9(2n+1)$

$\sum_{k=1}^n \frac{108k^3}{n^3} = \frac{108}{n^3} \sum_{k=1}^n k^3 = \frac{108}{n^3} \frac{n^2(n+1)^2}{4} = 27(n+1)^2$

$\sum_{k=1}^n \frac{81k^4}{n^4} = \frac{81}{n^4} \sum_{k=1}^n k^4 = \frac{81}{n^4} \frac{n^3(n+1)^3}{30} = \frac{27}{10}(n+1)^3$

Combining the Terms

Now, we can combine the terms:

$\sum_{k=1}^n (1 + \frac{12k}{n} + \frac{54k^2}{n^2} + \frac{108k^3}{n^3} + \frac{81k^4}{n^4}) = n + 6(n+1) + 9(2n+1) + 27(n+1)^2 + \frac{27}{10}(n+1)^3$

Simplifying the Expression

We can simplify the expression by combining like terms:

$n + 6n + 6 + 18n + 9 + 27n^2 + 54n + 27n^2 + 27n + \frac{27}{10}n^3 + \frac{27}{10}n^2 + \frac{27}{10}n + \frac{27}{10}$

Combining Like Terms

Combining like terms, we get:

$27n^2 + \frac{27}{10}n^3 + 51n + 36 + \frac{27}{10}$

Simplifying the Expression

We can simplify the expression further by combining like terms:

$\frac{27}{10}n^3 + 27n^2 + 51n + \frac{372}{10}$

Evaluating the Limit

Now, we can evaluate the limit as $n$ approaches infinity:

$\lim_{n \rightarrow \infty} \frac{3}{n} (\frac{27}{10}n^3 + 27n^2 + 51n + \frac{372}{10})$

Simplifying the Expression

We can simplify the expression by dividing each term by $n$:

$\lim_{n \rightarrow \infty} \frac{3}{n} (\frac{27}{10}n^3 + 27n^2 + 51n + \frac{372}{10}) = \lim_{n \rightarrow \infty} (27n^2 + 27n + \frac{51}{n} + \frac{372}{10n})$

Evaluating the Limit

As $n$ approaches infinity, the terms $\frac{51}{n}$ and $\frac{372}{10n}$ approach 0:

$\lim_{n \rightarrow \infty} (27n^2 + 27n + \frac{51}{n} + \frac{372}{10n}) = \lim_{n \rightarrow \infty} (27n^2 + 27n)$

Simplifying the Expression

We can simplify the expression by combining like terms:

$\lim_{n \rightarrow \infty} (27n^2 + 27n) = \lim_{n \rightarrow \infty} 27n^2$

Evaluating the Limit

As $n$ approaches infinity, the term $27n^2$ approaches infinity:

$\lim_{n \rightarrow \infty} 27n^2 = \infty$

Conclusion

In this article, we evaluated the limit of a summation as $n$ approaches infinity. The limit was $\lim_{n \rightarrow \infty} \frac{3}{n} \sum_{k=1}^n\left(2+\frac{3 k}{n}\right)^4$. We simplified the expression using the binomial theorem and evaluated the limit as $n$ approaches infinity. The final answer was $\infty$.

Introduction

In our previous article, we evaluated the limit of a summation as $n$ approaches infinity. The limit was $\lim_{n \rightarrow \infty} \frac{3}{n} \sum_{k=1}^n\left(2+\frac{3 k}{n}\right)^4$. In this article, we will answer some common questions related to the evaluation of this limit.

Q: What is the main concept behind evaluating the limit of a summation?

A: The main concept behind evaluating the limit of a summation is to simplify the expression using mathematical techniques such as the binomial theorem and then evaluate the limit as $n$ approaches infinity.

Q: How do you apply the binomial theorem to evaluate the limit of a summation?

A: To apply the binomial theorem, you need to expand the polynomial expression inside the summation using the binomial theorem formula. Then, you can simplify the expression by combining like terms and evaluate the limit as $n$ approaches infinity.

Q: What is the significance of the term $\frac{3}{n}$ in the limit expression?

A: The term $\frac{3}{n}$ is a scaling factor that affects the value of the limit. As $n$ approaches infinity, the term $\frac{3}{n}$ approaches 0, which means that the limit is dominated by the term inside the summation.

Q: How do you handle the term $\frac{51}{n}$ in the limit expression?

A: As $n$ approaches infinity, the term $\frac{51}{n}$ approaches 0. This means that the term $\frac{51}{n}$ has a negligible effect on the value of the limit.

Q: What is the final answer to the limit expression?

A: The final answer to the limit expression is $\infty$. This means that the limit approaches infinity as $n$ approaches infinity.

Q: Can you provide a step-by-step solution to the limit expression?

A: Yes, we can provide a step-by-step solution to the limit expression. Here is the solution:

1. Expand the polynomial expression inside the summation using the binomial theorem.
2. Simplify the expression by combining like terms.
3. Evaluate the limit as $n$ approaches infinity.
4. Simplify the expression further by dividing each term by $n$.
5. Evaluate the limit as $n$ approaches infinity.

Q: What are some common mistakes to avoid when evaluating the limit of a summation?

A: Some common mistakes to avoid when evaluating the limit of a summation include:

• Not expanding the polynomial expression inside the summation using the binomial theorem.
• Not simplifying the expression by combining like terms.
• Not evaluating the limit as $n$ approaches infinity.
• Not simplifying the expression further by dividing each term by $n$.

Q: Can you provide some examples of limits of summations?

A: Yes, here are some examples of limits of summations:

• $\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$
• $\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n k^3 = \frac{n^2(n+1)^2}{4}$
• $\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n k^4 = \frac{n^3(n+1)^3}{30}$

Conclusion

In this article, we answered some common questions related to the evaluation of the limit of a summation. We provided a step-by-step solution to the limit expression and discussed some common mistakes to avoid when evaluating the limit of a summation. We also provided some examples of limits of summations.