Q.

For any increasing sequence of real numbers a1,a2,a3,... let us denote A as a1,a2,a3,a4,..., and a sequence ΔA is defined as a2a1,a3a2,a4a3,........ Suppose that all of the terms of the sequence Δ(ΔA) are 1 and that a19=a92=0. Then the sum of digit(s) of a3 is M, then the value of [M2] is ____, where [.] is greatest integer function.

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answer is 5.

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Detailed Solution

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Suppose that the first term of the sequence ΔA is d then
ΔA={d,d+1,d+2,.......(d+(n1))} 
Hence A=(a1,a1+d,a1+d+(d+1)),  
a1+d+(d+1)+(d+2)...........)
an=a1+(n1)d+12(n1)(n2) 
so an is a quadratic polynomial in n.
so an=(nα)(nβ)2   
Since a19=a92=0 we must have an=12(n19)(n92)  
So a3=12(319)(392)=712 

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