<s>[INST] Finding second last digit of term in recursive series A sequence of numbers $a_{1}, a_{2}, a_{3},...$ is generated by the formula $a_{1}=3, a_{n+1}=a_{n} + a_{n}^{2}$ for integers $n \geqslant 1$. What is the second last digit of the number $a_{1993}$?
I am not able to recognise a pattern in the last two digits when trying out the recursive formula for the first five iterations. How should this question be approached?
 [/INST] Since we have to find the second last digit of $a_{1993}$ , it is enough to calculate the remainder $a_{1993}$ leaves upon division by $100$.
Note that $a_{n+1}\equiv a_n +a_{n}^2\pmod {100}$
Note that , $a_1=3$ and $a_{1}^2=9$ and $a_2=2$. Using this we can write $$a_2\equiv12\pmod{100}$$
$$a_{2}^2\equiv 144\equiv 44\pmod{100}$$
$$a_3\equiv a_2+a_{2}^2\equiv 56\pmod{100}$$
$$a_{3}\equiv 56\pmod{100}\implies a_{3}^2\equiv 3136\equiv36\pmod{100}$$
$$a_4\equiv a_3+a_{3}^2\equiv 92\pmod{100}$$
$$a_4\equiv 92\pmod{100}\implies a_{4}^2\equiv 8464\equiv 64\pmod{100}$$
$$a_5\equiv a_4+a_{4}^2\equiv 92+64\equiv 56\pmod{100}$$
Similarly, $a_6\equiv 92\pmod{100}$ ,$a_7\equiv 56\pmod{100}$ and so on.......
This pattern of $56$ and $92$ repeats indefinitely . So , $a_{1993}\equiv 56$ or $92\pmod {100}$. I leave it upto you to find out whether the second last digit is $5$ or $9$.
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