Math Problem Statement

I have a series where the numbers go 2+5+9+14+20+27+...

Find an explicit formula for this infinite series.

Solution

The series you provided follows a pattern where the numbers increase in a regular manner. Let’s analyze this to find an explicit formula.

Given sequence:

2,5,9,14,20,27,2, 5, 9, 14, 20, 27, \dots

Step 1: Identify the differences between consecutive terms.

52=35 - 2 = 3 95=49 - 5 = 4 149=514 - 9 = 5 2014=620 - 14 = 6 2720=727 - 20 = 7

The differences between consecutive terms are increasing by 1 each time (3, 4, 5, 6, 7,...). This suggests the sequence is quadratic.

Step 2: Assume a quadratic form for the general term.

A quadratic sequence can be represented by: an=An2+Bn+Ca_n = An^2 + Bn + C where ana_n is the nth term of the sequence, and AA, BB, and CC are constants we need to determine.

Step 3: Use known terms to find AA, BB, and CC.

We have:

  • a1=2a_1 = 2
  • a2=5a_2 = 5
  • a3=9a_3 = 9

Now, substitute these values into the quadratic equation.

For n=1n = 1: A(1)2+B(1)+C=2A+B+C=2(Equation 1)A(1)^2 + B(1) + C = 2 \quad \Rightarrow A + B + C = 2 \quad \text{(Equation 1)}

For n=2n = 2: A(2)2+B(2)+C=54A+2B+C=5(Equation 2)A(2)^2 + B(2) + C = 5 \quad \Rightarrow 4A + 2B + C = 5 \quad \text{(Equation 2)}

For n=3n = 3: A(3)2+B(3)+C=99A+3B+C=9(Equation 3)A(3)^2 + B(3) + C = 9 \quad \Rightarrow 9A + 3B + C = 9 \quad \text{(Equation 3)}

Step 4: Solve the system of equations.

We now have the system of equations:

  1. A+B+C=2A + B + C = 2
  2. 4A+2B+C=54A + 2B + C = 5
  3. 9A+3B+C=99A + 3B + C = 9

Subtract Equation 1 from Equation 2: (4A+2B+C)(A+B+C)=52(4A + 2B + C) - (A + B + C) = 5 - 2 3A+B=3(Equation 4)3A + B = 3 \quad \text{(Equation 4)}

Subtract Equation 2 from Equation 3: (9A+3B+C)(4A+2B+C)=95(9A + 3B + C) - (4A + 2B + C) = 9 - 5 5A+B=4(Equation 5)5A + B = 4 \quad \text{(Equation 5)}

Now subtract Equation 4 from Equation 5: (5A+B)(3A+B)=43(5A + B) - (3A + B) = 4 - 3 2A=1A=122A = 1 \quad \Rightarrow A = \frac{1}{2}

Substitute A=12A = \frac{1}{2} into Equation 4: 3(12)+B=332+B=3B=323\left(\frac{1}{2}\right) + B = 3 \quad \Rightarrow \frac{3}{2} + B = 3 \quad \Rightarrow B = \frac{3}{2}

Now substitute A=12A = \frac{1}{2} and B=32B = \frac{3}{2} into Equation 1: 12+32+C=22+C=2C=0\frac{1}{2} + \frac{3}{2} + C = 2 \quad \Rightarrow 2 + C = 2 \quad \Rightarrow C = 0

Step 5: Write the explicit formula.

The explicit formula for the nth term of the sequence is: an=12n2+32na_n = \frac{1}{2}n^2 + \frac{3}{2}n

This is the formula for the given sequence.


Would you like more details on how this formula works? Or do you have any other questions?

Here are some related questions:

  1. How can I recognize the type of sequence (linear, quadratic, etc.) from the pattern?
  2. How can I use this formula to find the 10th term of the sequence?
  3. What happens if the sequence were cubic? How would the formula change?
  4. Can this method be used for sequences with more complex patterns, like alternating terms?
  5. How do I find the sum of the first nn terms of this sequence?

Tip: Always check the differences between terms first—it helps you determine whether the sequence is arithmetic, quadratic, or more complex!

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Math Problem Analysis

Mathematical Concepts

Sequences
Quadratic Sequences
Algebra
Difference of Terms

Formulas

a_n = An^2 + Bn + C

Theorems

Method of solving quadratic equations using system of equations

Suitable Grade Level

Grades 7-9