GATE | GATE-CS-2016 (Set 1) | Question 37

Consider the recurrence relation a1 = 8, an = 6n2 + 2n + an-1. Let a99 = k x 104. The value of K is _____

 
Note : This question was asked as Numerical Answer Type.

(A) 190
(B) 296
(C) 198
(D) 200


Answer: (C)

Explanation: a1 = 8
an = 6n2 + 2n + an-1
 
an = 6[n2 + (n-1)2] + 2[n + (n-1)] + an-2
 
Continuing the same way till n=2, we get
an = 6[n2 + (n-1)2 + (n-2)2 + … + (2)2] + 2[n + (n-1) + (n-2) + … + (2)] + a1
 
an = 6[n2 + (n-1)2 + (n-2)2 + … + (2)2] + 2[n + (n-1) + (n-2) + … + (2)] + 8
 
an = 6[n2 + (n-1)2 + (n-2)2 + … + (2)2] + 2[n + (n-1) + (n-2) + … + (2)] + 6 + 2
 
an = 6[n2 + (n-1)2 + (n-2)2 + … + (2)2 + 1] + 2[n + (n-1) + (n-2) + … + (2) + 1]
 
an = (n)*(n+1)*(2n+1) + (n)(n+1) = (n)*(n+1)*(2n+2)
 
an = 2*(n)*(n+1)*(n+1) = 2*(n)*(n+1)2
 
Now, put n=99.
a99 = 2*(99)*(100)2 = 1980000 = K * 104
Therefore, K = 198.
 
Thus, C is the correct choice.

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