Question Number 22423 by Tinkutara last updated on 17/Oct/17 $$\mathrm{Prove}\:\mathrm{that}\:\mathrm{no}\:\mathrm{three}\:\mathrm{consecutive} \\ $$$$\mathrm{binomial}\:\mathrm{coefficient}\:\mathrm{can}\:\mathrm{be}\:\mathrm{in}\:\mathrm{G}.\mathrm{P}.\:\mathrm{or}\:\mathrm{H}.\mathrm{P}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 22392 by Tinkutara last updated on 17/Oct/17 $$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{odd}\:\mathrm{coefficients} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{expansion}\:\mathrm{of}\:\left(\mathrm{1}\:+\:\mathrm{2}{x}\:−\:\mathrm{3}{x}^{\mathrm{2}} \right)^{\mathrm{1025}} \\ $$$$\mathrm{is}\:\mathrm{an}\:\mathrm{even}\:\mathrm{integer}. \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 22394 by Tinkutara last updated on 17/Oct/17 $$\mathrm{Prove}\:\mathrm{that}\::\:\frac{\:^{{n}} {C}_{\mathrm{0}} }{{n}}−\frac{\:^{{n}} {C}_{\mathrm{1}} }{{n}+\mathrm{1}}+\frac{\:^{{n}} {C}_{\mathrm{2}} }{{n}+\mathrm{2}}−…+\left(−\mathrm{1}\right)^{{n}} .\frac{\:^{{n}} {C}_{{n}} }{\mathrm{2}{n}}=\frac{\mathrm{1}}{{n}.^{\mathrm{2}{n}} {C}_{{n}} } \\ $$ Terms of…
Question Number 22389 by Tinkutara last updated on 17/Oct/17 $${Simplify}: \\ $$$$\:^{{n}−\mathrm{1}} {C}_{\mathrm{2}} +\mathrm{2}\:^{{n}−\mathrm{2}} {C}_{\mathrm{2}} +\mathrm{3}\:^{{n}−\mathrm{3}} {C}_{\mathrm{2}} +…+\left({n}−\mathrm{2}\right)\:^{\mathrm{2}} {C}_{\mathrm{2}} \\ $$ Terms of Service Privacy…
Question Number 22379 by Tinkutara last updated on 16/Oct/17 $$\mathrm{For}\:\mathrm{each}\:\mathrm{positive}\:\mathrm{integer}\:{n},\:\mathrm{define}\:{a}_{{n}} \:= \\ $$$$\mathrm{20}\:+\:{n}^{\mathrm{2}} ,\:\mathrm{and}\:{d}_{{n}} \:=\:{gcd}\left({a}_{{n}} ,\:{a}_{{n}+\mathrm{1}} \right).\:\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{all}\:\mathrm{values}\:\mathrm{that}\:\mathrm{are}\:\mathrm{taken}\:\mathrm{by} \\ $$$${d}_{{n}} \:\mathrm{and}\:\mathrm{show}\:\mathrm{by}\:\mathrm{examples}\:\mathrm{that}\:\mathrm{each}\:\mathrm{of} \\ $$$$\mathrm{these}\:\mathrm{values}\:\mathrm{are}\:\mathrm{attained}. \\…
Question Number 153450 by liberty last updated on 07/Sep/21 $$\begin{cases}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} ={x}+{y}+\mathrm{2}}\\{\left({y}+\mathrm{1}\right)^{\mathrm{2}} ={y}+{z}+\mathrm{2}}\\{\left({z}+\mathrm{1}\right)^{\mathrm{2}} ={z}+{x}+\mathrm{2}}\end{cases} \\ $$ Answered by EDWIN88 last updated on 07/Sep/21 Commented by Tawa11…
Question Number 153446 by mathdanisur last updated on 07/Sep/21 $$\mathrm{L}^{−\mathrm{1}} \left\{\frac{\mathrm{s}}{\mathrm{s}^{\mathrm{2}} \:-\:\mathrm{12s}\:+\:\mathrm{40}}\right\}\:=\:? \\ $$ Commented by alisiao last updated on 07/Sep/21 $$=\:{L}^{−\mathrm{1}} \:\left\{\frac{\left({s}−\mathrm{6}\right)}{\left({s}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{4}}\:+\:\frac{\mathrm{6}}{\left({s}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{4}}\right\}…
Question Number 87886 by TawaTawa1 last updated on 06/Apr/20 Commented by jagoll last updated on 07/Apr/20 $$\mathrm{x}^{\mathrm{2}} −\mathrm{4x}+\mathrm{1}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{x}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{x}\:=\:\mathrm{2}\:\pm\:\sqrt{\mathrm{3}}\: \\ $$$$\mathrm{y}^{\mathrm{2}}…
Question Number 153412 by mathdanisur last updated on 07/Sep/21 $$\mathrm{If}\:\:\:\mathrm{a};\mathrm{b};\mathrm{c};\mathrm{d}\in\left(\mathrm{0};\infty\right)\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{a}^{\mathrm{3}} }{\mathrm{bc}}\:+\:\frac{\mathrm{b}^{\mathrm{3}} }{\mathrm{cd}}\:+\:\frac{\mathrm{c}^{\mathrm{3}} }{\mathrm{da}}\:+\:\frac{\mathrm{d}^{\mathrm{3}} }{\mathrm{ab}}\:\geqslant\:\mathrm{a}\:+\:\mathrm{b}\:+\:\mathrm{c}\:+\:\mathrm{d} \\ $$ Answered by puissant last updated on 07/Sep/21…
Question Number 153407 by liberty last updated on 07/Sep/21 Answered by Rasheed.Sindhi last updated on 08/Sep/21 $${x}\equiv\mathrm{2}\:\left({mod}\:\mathrm{5}\right)……..\left({i}\right) \\ $$$${x}\equiv\mathrm{1}\:\left({mod}\:\mathrm{3}\right)……..\left({ii}\right) \\ $$$${x}\equiv\mathrm{6}\:\left({mod}\:\mathrm{14}\right)……\left({iii}\right) \\ $$$$\underset{\underset{} {−}} {{x}\equiv\mathrm{5}\underline{\:\left({mod}\:\mathrm{11}\right)…….\left({iv}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:}\:\:\:\:\:\:\:\:\:}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…