Use f(0)=−3\displaystyle {f{{\left({0}\right)}}}=-{3}f(0)=−3 and f(1)=−1\displaystyle {f{{\left({1}\right)}}}=-{1}f(1)=−1 to compute f(4)\displaystyle {f{{\left({4}\right)}}}f(4) given:
f(p+2)=1f(p)+1f(p+1)\displaystyle {f{{\left({p}+{2}\right)}}}=\frac{{1}}{{f{{\left({p}\right)}}}}+\frac{{1}}{{f{{\left({p}+{1}\right)}}}}f(p+2)=f(p)1+f(p+1)1
f(4)=\displaystyle {f{{\left({4}\right)}}}=f(4)= Preview Question 6
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