Let #a_n# be a sequence given by: #{1, 6, 15, 28, 45,66,..., f(n)}#. Show that the generating function #f(n)# is of the form # an^2 + bn + c#. Find the formula by computing the coefficients #a, b, c#?

1 Answer

Answer:

#:. P_n^6 = 2n^2-n#

Explanation:

Strategy:
Take the given sequence find the difference between consecutive numbers:
#P_n = {1,6,15,28,45,66, 91,120,cdots}#
Step 1 #rArr# Layer 1
#{1,5,9,13,17,21, cdots}#
Step 2 #rArr# Layer 2 , Do it again
#{4, 4, 4, 4, 4,cdots}#
Taking the difference is in discrete math is the same as taking the derivative (i.e. slope). took two subtraction (two layers) before we reached a comstant number #4#, that means the sequence is polynomial growth.
Give that I asert that: #P_n = an^2+bn+c#
All I have to do now find the value of #a, b and c#
To solve for #a,b and c# I use the first 3 entry of the sequence setting #n = {1,2,3}#

#Eq.1 rArr##P_1 = a+b+c = 1#
#Eq.2 rArr##P_2 = 4a+2b+c = 6#
#Eq.3 rArr##P_3 = 9a+3b+c = 15#
#[[1,1,1], [4,2,1], [9,3,1]]xx[[a],[b],[c]] = [[1],[6],[15]]#

Solving a, b, c using any matrix calculator on the internet:
#[[a],[b],[c]] = [[2],[-1],[0]] #
#:. P_n^6 = 2n^2-n#

Check: #P_1^6 = 1; P_2^6 = 6; P_3^6 = 15;# checks out

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