How do you find the nth term of the sequence #2,5,10,17,26,37,...#?

Question

Luke Alvarez · Accepted Answer

# n^2+1 #

This is the GCSE approach:

Take the difference between consecutive terms, and then take the difference between those terms, as follows:

# {: ("Seq: ",2,,5,,10,,17,,26,,37,), ("1st Difference:",,3,,5,,7,,9,,11,,),("2nd difference:",,,2,,2,,2,,2,,,) :} #

So as the 2nd difference is constant, we can infer that the terms form a quadratic sequence, of the form:

# {u_n} = an^2 + bx + c #

and the quadratic coefficient is twice the 2nd difference, ie:

# 2a=2 => a=1 #

We now form a table where we take the original sequence and subtract the quadratic term to give a residue:

# {: (n:, 1,2,3,4,5,6), ("Seq ("n"): ", 2,5,10,17,26,37), (an^2=n^2: ,1,4,9,16,25,36), ("Residue:" ,1,1,1,1,1,1) :}#

So we can now also conclude that the linear part of the sequence (#bx+c#) form the terms #{1,1,1,1, ... }#, so we can conclude that #b=0# and #c=1# and hence the general term of the sequence is:

# \ \ \ \ \ u_n = n^2 + 0n + 1 # # :. u_n =n^2 + 1 #

Which we can verify as follows:

# {: (n:, 1,2,3,4,5,6), (n^2: ,1,4,9,16,25,36), (n^2+1 ,2,5,10,17,26,37) :}#

Isaiah Allen · Answer

the nth term is #n^2+1# The sequence above is quadratic, and we use the expression #an^2+bn+c# where 'a' represents #1/2# the second difference and 'c' is the 0th term. We then substitute numbers into the equation to find the value of 'b'. For example, the second difference in your sequence would be 2 because the 1 first differences are 3, 5, 7, 9 and 11. Thus 'a' = 1 because #2/2 = 1#. Then we work backwards to find 'c'. The first difference for the first two terms is 3. #3-2 = 1#. Then the first term - 1 = 1. After this we just substitute numbers into the equation to find the value of 'b'. #1n^2+1b+1=2#, when n = 1 #2+b=2# #b=0# so the nth term is #n^2 + 1#

Charles Anctil · Answer

undefined The nth term of the sequence $2, 5, 10, 17, 26, 37, \ldots$ can be found using the formula $a_n = n^2 + 1$.

Isaiah Allen · Answer

undefined To find the nth term of the sequence 2, 5, 10, 17, 26, 37,..., we observe that the differences between consecutive terms are increasing by 3 each time. This suggests that the sequence is generated by a quadratic expression.

The differences between consecutive terms are:
3, 5, 7, 9, 11,...

These differences increase by 2 each time. This indicates that the second differences are constant, suggesting a quadratic sequence.

To find the nth term of a quadratic sequence, we start by finding the formula for the nth term in terms of n. We can represent the nth term as an expression in the form $ an^2 + bn + c $, where a, b, and c are constants to be determined.

We can find the values of a, b, and c by substituting the values of n and the corresponding terms in the sequence into the general expression for the nth term.

Using the first three terms of the sequence:
When n = 1, term = 2
When n = 2, term = 5
When n = 3, term = 10

Substituting these values into the expression $ an^2 + bn + c $, we can form a system of equations and solve for a, b, and c.

$$a(1)^2 + b(1) + c = 2$$
$$a(2)^2 + b(2) + c = 5$$
$$a(3)^2 + b(3) + c = 10$$

Solving this system of equations will give us the values of a, b, and c. Once we have these values, we can write the formula for the nth term of the sequence.