We have a peculiar series in which a square is written first and then the number of consecutive numbers that follow the square (including the square) is equal to the square divided by its square root.
If the series starts from 1 as 1, 4, 5, 9, 10, 11, 16, 17, 18, 19, ... , then what is the 2016th number in the series?
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Lets divide the sequence into segments as follows:-
1 ∣ 4 , 5 ∣ 9 , 1 0 , 1 1 ∣ 1 6 , 1 7 , 1 8 , 1 9 ⋯ .
So, 1 ,is the first term in the sequence.
4 is the third term.
9 is the sixth term.
1 6 is the tenth term.
So, n 2 is the ( ( i = 1 ∑ n i ) − ( n − 1 ) ) th term of sequence.
So, let x 2 be the first square in the 2 0 1 6 th segment.
Then i = 1 ∑ x i = 2 0 1 6 2 x ( x + 1 ) = 2 0 1 6 x 2 + x − 4 0 3 2 = 0 x 2 + 6 4 x − 6 3 x − 4 0 3 2 = 0 ( x + 6 4 ) ( x − 6 3 ) = 0 x = − 6 4 ; x = 6 3 .
But, we neglect x = − 6 4 , because i = 1 ∑ − 6 4 i does not exist.
So, x = 6 3 .
So, the perfect square of the 2 0 1 6 th segment is 6 3 2 = 3 9 6 9 .
So, the 2 0 1 6 th term of the sequence is 3 9 7 9 + 6 2 = 4 0 3 1