claireroche1
contestada

What is the nth term of the sequence: 3,11,25,45
I know it begins with 3n^2
I entered 3n^2-(n-1) and it marked it as incorrect? Please help :)

Respuesta :

bartdrinksmalk
This solution might be a bit out there, but it should probably work.

For this, we will consider sequences to have an invisible "0th" term to make calculation easier.

For the simple sequence {0}, 1, 3, 6, 10,... ({0}+1=1, 1+2=3, 3+3=6, 6+4=10,...), the general formula is
[tex]\frac{n(n+1)}{2}[/tex]
Note the bold number increases by 1 each time.

We have {number}, 3, 11, 25, 45,... (3+8=11, 11+14=25, 25+20=45,...)
The bold number increases by 6 each time.

So we consider next: {0}, 6*1, 6*3, 6*6, 6*10,... ({0}+6=6, 6+12=18, 18+18=36, 36+24=60,...) and we get
[tex]6*\frac{n(n+1)}{2}=3n(n+1)[/tex]

From the 1st term to the 2nd term, this sequence has an increase of 12.

But we want it to have an increase of 8, so we subtract 4 from each bold number.

({0}+2=2, 2+8=10, 10+14=24, 24+40=64,...)

From each bold number, we have subtracted 4, so each term decreases by [tex]4n[/tex].

We now have
[tex]3n(n+1)-4n[/tex]

But instead of {number}, 2, 10, 24, 44,..., we want {number}, 3, 11, 25, 45,... So we add 1 to each term and get
[tex]3n(n+1)-4n+1[/tex]

Now we expand this expression out:
[tex]3n(n+1)-4n+1=3n^2+3n-4n+1=3n^2-n+1[/tex]

So remove the parentheses and make the 1 positive.

The [tex]{n^{th}}[/tex] term of the sequence [tex]3,11,25,45, \ldots[/tex] is [tex]\boxed{3{n^2} - n + 1}.[/tex]

Further Explanation:

Given:

The sequence is [tex]3,11,25,45, \ldots.[/tex]

Explanation:

The given sequence is [tex]3,11,25,45, \ldots.[/tex]

The first term of the sequence is 3, second term of the sequence is 11, third term is 25 and the fourth term is 45.

The difference between the first and second term can be obtained as follows,

[tex]\begin{aligned}{\text{Difference}} &= 11 - 3\\&= 8\\\end{aligned}[/tex]

The difference between the second and third term is 14.

The difference between the third term and fourth term is 20.

The series can be expressed as follows,

[tex]{\text{Series}} = 3,3 + 8,11 + 14,25 + 20, \ldots[/tex]

The difference between the consecutive terms is increases by 6.

Consider the series as [tex]0,6 \times 1,{\text{ }}6 \times 3,{\text{ }}6 \times 6,{\text{ 6}} \times {\text{10,}} \ldots.[/tex]

Apply principal of mathematical induction.

[tex]6 \times \dfrac{{n\left( {n + 1} \right)}}{2} = 3n\left( {n + 1} \right)[/tex]

The [tex]{n^{th}}[/tex] term of the sequence can be obtained as follows,

[tex]\begin{aligned}{n^{th}}{\text{ term}} &= 3n\left( {n + 1} \right) - 4n + 1\\&= 3{n^2} + 3n - 4n + 1\\&= 3{n^2} - n + 1\\\end{aligned}[/tex]

The [tex]{n^{th}}[/tex] term of the sequence [tex]3,11,25,45[/tex], [tex]\ldots[/tex] is [tex]\boxed{3{n^2} - n + 1}.[/tex]

Learn more:

  1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.
  2. Learn more about equation of circle brainly.com/question/1506955.
  3. Learn more about range and domain of the function https://brainly.com/question/3412497

Answer details:

Grade: High School

Subject: Mathematics

Chapter:Principal of Mathematical Induction

Keywords: sequence, 3, 11, 25, 45, explicit rule, common ratio, first term, second term, sum of sequence, nth term sequence.

Otras preguntas