Last digit

A progression is defined by a 1 = 3 a_1=3 and a n + 1 = a n + a n 2 a_{n+1}=a_n+a_{n}^2 for every natural number n 1 n\geq 1 . What is the last digit of a 2015 a_{2015} ?


The answer is 6.

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1 solution

a1 = 3 a2 = 12 , a3 = 156 Now last digit of a4 will be 6²+6 = 2, Similarly, Last digit of a5 will be 2²+2 = 6 This cycle continues . since 2015 is an odd number last digit will be 6

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