Use binet's formula to find the 1995th term of the series hint: it's the fibonnacia series
so \(F(n)=\frac{\phi^n-(-\phi)^n}{\sqrt5}\)Let n=1995
\(F(1995)=\frac{\phi^{1995}-(-\phi)^{1995}}{\sqrt{5}}\)
Let Phi=1.6 \(F(1995)=23668326.9616766279790584\) divided by 8 it equals
\(=2958540.8702095784973823\)
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