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

##### 2 Answers

the nth term is

#### Explanation:

The sequence above is quadratic, and we use the expression

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

so the nth term is

#### Explanation:

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 (

# \ \ \ \ \ 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) :}#