How do you find the next three terms in 2,5,10,17,26,....?

2 Answers
George C.
Nov 10, 2015

Examine the sequence of differences of this sequence and its sequence of differences to find the next three terms, #37, 50, 65# and the general formula #a_n = n^2+1#

Explanation:

To find a pattern in this sequence, first write out the original sequence:

(i) #color(blue)(2),5,10,17,26#

Then write out the sequence of differences between successive terms of that sequence:

(ii) #color(blue)(3),5,7,9#

Then write out the sequence of differences of that sequence:

(iii) #color(blue)(2),2,2#

Having reached a constant sequence, there are a couple of things we can do:

(1) We can find the next three elements of the original sequence as requested.

To do this, add three more #2#'s to the last sequence:

#color(blue)(2),2,2,color(red)(2),color(red)(2),color(red)(2)#

Then add three more terms to the previous sequence using the three new elements of this sequence as differences:

#color(blue)(3),5,7,9,color(red)(11),color(red)(13),color(red)(15)#

Then add three more terms to the original sequence using the three new elements of this sequences as differences:

#color(blue)(2),5,10,17,26,color(red)(37),color(red)(50),color(red)(65)#

(2) We can find a general formula for then #n#th term #a_n# of the sequence by using the initial terms #color(blue)(2)#, #color(blue)(3)#, #color(blue)(2)# of the sequences (i), (ii) and (iii) as coefficients:

#a_n = color(blue)(2)/(0!) + color(blue)(3)/(1!) (n-1) + color(blue)(2)/(2!) (n-1)(n-2)#

#=2+3n-3+n^2-3n+2 = n^2+1#

EZ as pi
Apr 16, 2018

#2," "5," "10," "17," "26," "color(blue)(37," "50," "65)#

Explanation:

Looking at the sequence it should be clear that it is neither arithmetic nor geometric.

#2," "5," "10," "17," "26#

Look at the differences:

,#3," "5," "7," "9.....#

This pattern is a clue that squares are involved.

Look at the pattern of the differences:

#2," "2," "2," "2..." "larr# indicates that it is a pattern of squares.

#T_n rarr" "2," "5," "10," "17," "26#
#nrarr" "1," "2," "3," "4," "5#
#n^2rarr" "1," "4," "9," "16," "25#

Comparing the square numbers with the given sequence, we can see that:

#T_n = n^2+1#

#T_6 = 6^2+1=37#

and so on to get the next terms as

#2," "5," "10," "17," "26," "color(blue)(37," "50," "65)#

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