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?

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 $9$.
