Math Problem Statement

2.14 2; 6; 10 ... are the first differences of a quadratic pattern. 2.14.1 Determine the first difference between T_37 and T_38 of the quadratic pattern. 2.14.2 If the 19th term of a quadratic pattern is 651, prove that the general term will be given by T_n = 2n^2 - 4n + 5. 2.14.3 Prove that all terms of the quadratic pattern will be odd.

Solution

Let's break down each part of this question and solve it step by step:

2.14.1 First difference between T37T_{37} and T38T_{38}

The first differences provided are 2,6,10,2, 6, 10, \ldots, and these suggest that the differences are increasing by 4 each time, meaning the quadratic pattern follows a certain second difference pattern.

The general quadratic sequence has the form:

Tn=an2+bn+cT_n = an^2 + bn + c

The first differences between consecutive terms follow a linear pattern:

ΔTn=Tn+1Tn=2a(n)+b\Delta T_n = T_{n+1} - T_n = 2a(n) + b

Given that the first differences are 2,6,10,2, 6, 10, \ldots, it suggests that the differences follow an arithmetic sequence with a common difference of 4.

To find the first difference between T37T_{37} and T38T_{38}, we'll need the general term of the first difference, i.e.,

ΔTn=2a(n)+b\Delta T_n = 2a(n) + b

Since the first difference follows a linear pattern starting at 2 and increasing by 4, this implies:

ΔTn=4n2\Delta T_n = 4n - 2

Thus, the first difference between T37T_{37} and T38T_{38} is:

ΔT37=4(37)2=1482=146\Delta T_{37} = 4(37) - 2 = 148 - 2 = 146 ΔT38=4(38)2=1522=150\Delta T_{38} = 4(38) - 2 = 152 - 2 = 150

So, the first difference between T37T_{37} and T38T_{38} is 150 - 146 = 4.

2.14.2 Proof of the general term given the 19th term is 651

We are given that the general term of the quadratic pattern is Tn=2n24n+5T_n = 2n^2 - 4n + 5, and the 19th term T19=651T_{19} = 651.

Let's substitute n=19n = 19 into the given expression:

T19=2(19)24(19)+5T_{19} = 2(19)^2 - 4(19) + 5 T19=2(361)76+5=72276+5=651T_{19} = 2(361) - 76 + 5 = 722 - 76 + 5 = 651

Hence, the given general term Tn=2n24n+5T_n = 2n^2 - 4n + 5 is correct.

2.14.3 Prove that all terms of the quadratic pattern will be odd

The general term is Tn=2n24n+5T_n = 2n^2 - 4n + 5.

Let's analyze the expression modulo 2 (since we're interested in whether the terms are odd).

Tn=2n24n+5T_n = 2n^2 - 4n + 5

For any integer nn:

  • 2n22n^2 is always even.
  • 4n-4n is always even.
  • +5+5 is odd.

So, TnT_n simplifies to:

Tn=eveneven+odd=oddT_n = \text{even} - \text{even} + \text{odd} = \text{odd}

Thus, all terms of the quadratic pattern will indeed be odd.


Would you like more details or explanations on any of the steps?

Related Questions:

  1. What is the second difference for a quadratic sequence, and why is it constant?
  2. How do you derive the general term of a quadratic sequence from the first and second differences?
  3. Can you explain how to verify a general term by substituting values of nn?
  4. How do arithmetic progressions relate to first differences in sequences?
  5. Can you explore the relationship between even and odd terms in quadratic patterns?

Tip:

When working with quadratic sequences, always check the first and second differences to identify patterns and help determine the general formula.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Quadratic Patterns
Arithmetic Sequences
Modular Arithmetic

Formulas

T_n = an^2 + bn + c
First difference: ΔT_n = 2a(n) + b

Theorems

Properties of Quadratic Sequences
Arithmetic Progression Theorem

Suitable Grade Level

Grades 9-12