1a_{n+1}-a_n=\frac{a_n^2}{a_{n-1}}\\ \frac{a_{n+1}}{a_n}-\frac{a_n}{a_{n-1}}=1
from here its quite apparent that
a_n=n!
Let a_0 = a_1 = 1 and a_{n+1} = 1 + \frac{a_1^2}{a_0} + \frac{a_2^2}{a_1} + ...+ \frac{a_n^2}{a_{n-1}}
Find a closed formula for an (i.e. express an as a function of n)
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3 Answers
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341yeah good! Sincere thanks for writing out the solution instead of just calling out the answer :D