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Sequences IV

Sequences IV

Last week we discussed this problem which was given in the Intermediate category of the 3rd International Vedic Mathematics Olympiad (IVMO 2023):

“A sequence starts, 4, 5, 8, 13, 20,… What is the 98th term in this sequence?

A) 9511       B) 9413      C) 9697      D) 9707      E) 9805”

We also wrote that

“The given sequence is  :    4           5          8         13        20  

“The first differences are:       1        3          5         7

“The second differences are:       2         2          2

“The common second difference of 2 between the first differences confirm that the sequence is quadratic in the form of An^2 + Bn + C and A = 1.” (We will explain why A is half the common second difference in our next post about sequences)

“The first differences are increasing by 2, so the difference between the 1st term and the 0th term  of the sequence must also be 2  or 1 – 2 = -1. So, the 0th term must be [4 – (-1) = 5). Using this value in A(0)^2 + B(0) + C = 5,   we can see that C = 5.”

We will now explain how we got the value A = 1 for this particular sequence and then discuss another way of determining the values of the other coefficients.

We have mentioned that each term of the sequence can be represented by the quadratic expression An^2 + Bn + C. So, we have the first differences of the first 5 terms as:

[A(2)^2 + B(2) + C] –  [A(1)^ + B(1) + C] = [4A + 2B + C] – [A + B + C] = 3A + B

[A(3)^2 + B(3) + C] –  [A(2)^ + B(2) + C] = [9A + 3B + C] – [4A + 2B + C] = 5A + B

[A(4)^2 + B(4) + C] –  [A(3)^ + B(3) + C] = [16A + 4B + C] – [9A + 3B + C] = 7A + B

[A(5)^2 + B(5) + C] –  [A(4)^ + B(4) + C] = [25A + 5B + C] – [16A + 4B + C] = 9A + B

And the second differences are:

            [5A + B] – [3A + B] = 2A

[7A + B] – [5A + B] = 2A

            [9A + B] – [7A + B] = 2A

We can see that the common second difference is twice the value of A. In our example, the common second difference is 2 and therefore, the value of the coefficient A is 1.

Knowing A, we can easily remove the An^2  in each term of the sequence to reduce them to BN + C.  We now have

1^2 + 1B + C =  4; B + C = 3

2^2 + 2B + C = 5; 2B + C = 1

            [2B + C] – [B + C] = 1 – 3

            B = – 2

            Since B + C = 3; -2 + C = 3; C = 5

The nth term of the given sequence is n^2 – 2n + 5 and the 98th term is 98^2 – 2(98) + 5 or

B) 9413.

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