Finding The Nth Term Of The Sequence 2, 15, 45, 110, 210

by THE IDEN 57 views

Finding patterns in sequences is a fascinating area of mathematics. Let's dive into the sequence 2, 15, 45, 110, 210 and explore how to determine the formula for its nth term, which is given by n(n+1)(3n-1)/2. This article will provide a comprehensive explanation of the process, ensuring a clear understanding for readers of all backgrounds. We'll break down the problem step by step, starting with recognizing the type of sequence, employing difference methods, deriving the general formula, and validating it with the given terms. Understanding these methods is not only useful for this specific problem but also for tackling other sequence-related challenges in mathematics.

Identifying the Sequence Type

To begin, it's essential to identify what type of sequence we're dealing with. The sequence 2, 15, 45, 110, 210 doesn't immediately appear to be arithmetic (where the difference between consecutive terms is constant) or geometric (where the ratio between consecutive terms is constant). To confirm this, let's calculate the differences between consecutive terms. The first differences are 15 - 2 = 13, 45 - 15 = 30, 110 - 45 = 65, and 210 - 110 = 100. These differences are not constant, so it's not an arithmetic sequence. Furthermore, the ratios between consecutive terms (15/2, 45/15, 110/45, 210/110) are also not constant, ruling out a geometric sequence. This suggests that the sequence might be a polynomial sequence, where the nth term can be expressed as a polynomial function of n. To determine the degree of the polynomial, we can continue finding differences until we reach a constant difference. This method, known as the method of differences, is crucial for unraveling the underlying pattern of the sequence. The method of differences involves calculating the differences between successive terms, then calculating the differences between those differences, and so on, until a constant difference is obtained. The number of times we need to take the difference before we get a constant value gives us the degree of the polynomial that represents the sequence. This is a powerful technique for analyzing sequences that aren't straightforward arithmetic or geometric progressions.

The Method of Differences: Unveiling the Pattern

Let's apply the method of differences to our sequence: 2, 15, 45, 110, 210. We've already calculated the first differences: 13, 30, 65, 100. Now, let's find the second differences by subtracting consecutive first differences: 30 - 13 = 17, 65 - 30 = 35, and 100 - 65 = 35. These differences are still not constant. Next, we calculate the third differences: 35 - 17 = 18 and 35 - 35 = 0. Oops! It seems we have made a mistake in the previous calculation. Let's double-check the second differences: 30 - 13 = 17, 65 - 30 = 35, 100 - 65 = 35. These are correct. Now, the third differences: 35 - 17 = 18, 35 - 35 = 0. Hmmm, it still doesn't look right. Let's check the original sequence and the first differences again to ensure there are no errors. Original sequence: 2, 15, 45, 110, 210. First differences: 13, 30, 65, 100. These seem correct. Now the second differences: 17, 35, 35. Third differences: 35 - 17 = 18, 35 - 35 = 0. Okay, there's still a discrepancy. Let's extend the sequence by one more term to get a clearer picture. To do this, we need to hypothesize how the sequence might continue based on the observed pattern. However, without additional information, it's challenging to predict the next term with certainty. Let's step back and re-examine the differences we've calculated. It's crucial to ensure accuracy at each step, as a small error can propagate and lead to an incorrect conclusion. The method of differences is a powerful tool, but it requires careful attention to detail.

Correcting the Calculation and Finding the Constant Difference

Let's revisit the differences with extra care. First differences: 13, 30, 65, 100. Second differences: 17, 35, 35. Third differences: 35 - 17 = 18, 35 - 35 = 0. We made an error in assuming the third differences would become constant immediately. Let's calculate one more term in the second differences. To do this, we need to find the difference between the next two first differences. We don't have the next term in the original sequence, so we need to deduce a pattern. Looking at the first differences (13, 30, 65, 100), the differences between them (second differences) are increasing. Let's assume the next first difference would increase by a similar amount. The pattern in the first differences is increasing, but not linearly. However, the second differences give us more insight: 17, 35, 35. Let's assume the next second difference is also around 35. Then the next first difference would be approximately 100 + 35 = 135. So, the next term in the original sequence would be approximately 210 + 135 = 345. Now, we can calculate the next first difference: 345 - 210 = 135. The next second difference: 135 - 100 = 35. The third difference: 35 - 35 = 0. Still, the third difference is not constant. This suggests that the polynomial might be of a higher degree than initially expected. However, we are given the formula n(n+1)(3n-1)/2, which is a cubic polynomial (degree 3). This means we should expect to find constant differences at the third level. Let's reassess our calculations. Original sequence: 2, 15, 45, 110, 210. First differences: 13, 30, 65, 100. Second differences: 17, 35, 35. It seems there might be an error in the sequence itself or in the provided formula. Let's proceed with the method of differences assuming the given formula is correct and see if we can derive it. If we assume the formula is correct, the third differences should be constant. The fact that they are not suggests a potential issue with our calculations or the provided sequence terms. Let's try another approach.

Deriving the General Formula

Since the method of differences is proving challenging due to the non-constant third differences, let's try a different approach. We are given the formula n(n+1)(3n-1)/2, which we want to verify. This is a cubic polynomial, so we can express the general form of the nth term as an³ + bn² + cn + d, where a, b, c, and d are constants. We can use the first few terms of the sequence to create a system of equations and solve for these constants. For n = 1, the term is 2: a(1)³ + b(1)² + c(1) + d = 2. For n = 2, the term is 15: a(2)³ + b(2)² + c(2) + d = 15. For n = 3, the term is 45: a(3)³ + b(3)² + c(3) + d = 45. For n = 4, the term is 110: a(4)³ + b(4)² + c(4) + d = 110. This gives us the following system of equations:

  • a + b + c + d = 2
  • 8a + 4b + 2c + d = 15
  • 27a + 9b + 3c + d = 45
  • 64a + 16b + 4c + d = 110

Solving this system of equations will give us the values of a, b, c, and d. While this can be done manually using substitution or elimination, it's more efficient to use a matrix method or a calculator with equation-solving capabilities. Solving this system is crucial for confirming whether the given formula is indeed correct. The process involves a series of algebraic manipulations to isolate each variable and find its value. Once we have the values of a, b, c, and d, we can substitute them back into the general cubic form and compare the resulting polynomial with the given formula. If they match, it confirms the correctness of the formula; if not, it indicates a discrepancy that needs further investigation. This method highlights the connection between sequences and polynomials, demonstrating how algebraic techniques can be used to analyze and understand number patterns.

Solving the System of Equations

Let's solve the system of equations we derived:

  • a + b + c + d = 2
  • 8a + 4b + 2c + d = 15
  • 27a + 9b + 3c + d = 45
  • 64a + 16b + 4c + d = 110

Subtracting the first equation from the second, the second from the third, and the third from the fourth, we get:

  • 7a + 3b + c = 13
  • 19a + 5b + c = 30
  • 37a + 7b + c = 65

Subtracting the first of these new equations from the second and the second from the third, we get:

  • 12a + 2b = 17
  • 18a + 2b = 35

Subtracting the first of these equations from the second, we get:

  • 6a = 18

So, a = 3. Substituting a = 3 into 12a + 2b = 17, we get:

  • 12(3) + 2b = 17
  • 36 + 2b = 17
  • 2b = -19

So, b = -19/2. Substituting a = 3 and b = -19/2 into 7a + 3b + c = 13, we get:

  • 7(3) + 3(-19/2) + c = 13
  • 21 - 57/2 + c = 13
  • c = 13 - 21 + 57/2
  • c = -8 + 57/2
  • c = (-16 + 57)/2
  • c = 41/2

Substituting a = 3, b = -19/2, and c = 41/2 into a + b + c + d = 2, we get:

  • 3 - 19/2 + 41/2 + d = 2
  • (6 - 19 + 41)/2 + d = 2
  • 28/2 + d = 2
  • 14 + d = 2
  • d = -12

Thus, the general form is 3n³ - (19/2)n² + (41/2)n - 12. Multiplying by 2 to eliminate fractions, we have 6n³ - 19n² + 41n - 24. This polynomial appears quite different from the given formula, n(n+1)(3n-1)/2. Let's expand the given formula to compare: n(n+1)(3n-1)/2 = n(3n² + 2n - 1)/2 = (3n³ + 2n² - n)/2. Multiplying by 2, we get 3n³ + 2n² - n. There is a clear discrepancy between the derived polynomial and the expanded given formula. This strongly suggests an error in our calculations or a mistake in the original sequence terms. It's crucial to double-check all calculations and the given sequence to identify the source of the inconsistency.

Validating the Formula

To validate the given formula, n(n+1)(3n-1)/2, we will substitute the first few values of n and see if they match the terms in the sequence 2, 15, 45, 110, 210. For n = 1: (1(1+1)(3(1)-1))/2 = (1(2)(2))/2 = 2. This matches the first term. For n = 2: (2(2+1)(3(2)-1))/2 = (2(3)(5))/2 = 15. This matches the second term. For n = 3: (3(3+1)(3(3)-1))/2 = (3(4)(8))/2 = 48. This does not match the third term, which is 45. This indicates an error in the given formula or the sequence itself. Let's continue the validation to confirm. For n = 4: (4(4+1)(3(4)-1))/2 = (4(5)(11))/2 = 110. This matches the fourth term. For n = 5: (5(5+1)(3(5)-1))/2 = (5(6)(14))/2 = 210. This matches the fifth term. The formula seems to work for n = 1, 2, 4, and 5 but fails for n = 3. This is a crucial finding. It suggests that there might be a typo in the sequence or the formula might be a piecewise function that behaves differently for n = 3. Given the discrepancy, it's essential to carefully re-examine the sequence and the formula to identify the root cause. The fact that the formula works for most terms but not all highlights the importance of thorough validation in mathematical problem-solving.

Identifying the Error and Correcting the Sequence

The validation process revealed that the formula n(n+1)(3n-1)/2 does not produce the sequence 2, 15, 45, 110, 210 accurately, specifically for the third term. The formula yields 48 for n = 3, while the sequence provides 45. This discrepancy indicates a potential error either in the sequence itself or in the problem statement. To pinpoint the error, let's re-examine the method of differences and our calculations. We know the formula is a cubic polynomial, so the third differences should be constant if the sequence were correct. However, our earlier attempt to find constant third differences was unsuccessful, further suggesting an issue. If we assume the formula is correct, then the third term in the sequence should be 48, not 45. Let's replace 45 with 48 and see if the sequence now fits the formula and yields constant third differences. The corrected sequence would be 2, 15, 48, 110, 210. Recalculating the differences: First differences: 13, 33, 62, 100. Second differences: 20, 29, 38. Third differences: 9, 9. With this correction, we obtain constant third differences, supporting the hypothesis that the original sequence had an error. This process demonstrates the power of validation and error analysis in mathematical problem-solving. Identifying and correcting errors is a crucial part of the mathematical process, often leading to a deeper understanding of the underlying concepts.

Final Verification and Conclusion

With the corrected sequence 2, 15, 48, 110, 210, we can confidently say that the nth term of the sequence is indeed given by the formula n(n+1)(3n-1)/2. We arrived at this conclusion through a combination of methods: initially attempting the method of differences, then deriving a general cubic polynomial, and finally validating the given formula. The validation process was crucial in identifying an error in the original sequence, which, when corrected, allowed us to confirm the formula. This problem-solving journey highlights the importance of using multiple approaches and the necessity of careful validation in mathematics. By understanding these techniques, you can approach similar sequence-related problems with confidence. Remember, mathematics is not just about finding the right answer; it's about the process of exploration, discovery, and critical thinking. This particular problem illustrates how errors can be valuable learning opportunities, prompting a deeper analysis and a more thorough understanding of the underlying concepts.