At the given condition:

escumantsu

Answered 2021-07-20
Author has **18662** answers

asked 2021-02-09

asked 2021-01-19

Prove that

\(\displaystyle{F}_{{n}}=\frac{1}{{4}}{\left({F}_{{{n}-{2}}}+{F}_{{n}}+{F}_{{{n}+{2}}}\right)}\)

asked 2021-07-30

asked 2021-07-18

Prove the following using propositional logic:

asked 2020-12-31

asked 2021-02-01

asked 2021-07-20

\(\displaystyle{f}_{{n}}={10}\cdot{f}_{{{n}-{1}}}-{25}\cdot{f}_{{{n}-{2}}}\)

\(\displaystyle{f}_{{n}}=\alpha_{{1}}{\left(-{5}\right)}^{{n}}+\alpha_{{2}}{\left(-{5}\right)}^{{n}}\)

\(\displaystyle{f}_{{n}}=\alpha_{{1}}{\left(-{5}\right)}^{{n}}+\alpha_{{2}}{n}{\left(-{5}\right)}^{{n}}\)

\(\displaystyle{f}_{{n}}=\alpha_{{1}}{\left({5}\right)}^{{n}}+\alpha_{{2}}{n}{\left(-{5}\right)}^{{n}}\)

\(\displaystyle{f}_{{n}}=\alpha_{{1}}{\left({5}\right)}^{{n}}+\alpha_{{2}}{\left({5}\right)}^{{n}}\)

\(\displaystyle{f}_{{n}}=\alpha_{{1}}{\left({5}\right)}^{{n}}+\alpha_{{2}}{\left(-{5}\right)}^{{n}}\)

\(\displaystyle{f}_{{n}}=\alpha_{{1}}{\left({5}\right)}^{{n}}+\alpha_{{2}}{n}{\left({5}\right)}^{{n}}\)