Author: Tinku Tara

Question Number 71550 by mhmd last updated on 17/Oct/19 Answered by MJS last updated on 17/Oct/19 $$\mathrm{it}'\:\mathrm{wrong},\:\mathrm{it}\:\mathrm{must}\:\mathrm{be} \\ $$$$\begin{pmatrix}{{n}+\mathrm{1}}\\{{k}}\end{pmatrix}\:=\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\:+\begin{pmatrix}{{n}}\\{{k}−\mathrm{1}}\end{pmatrix} \\ $$$$\begin{pmatrix}{{n}}\\{{k}}\end{pmatrix}\:+\begin{pmatrix}{{n}}\\{{k}−\mathrm{1}}\end{pmatrix}\:=\frac{{n}!}{{k}!\left({n}−{k}\right)!}+\frac{{n}!}{\left({k}−\mathrm{1}\right)!\left({n}−{k}+\mathrm{1}\right)!}= \\ $$$$=\frac{{n}!\left({n}−{k}+\mathrm{1}\right)}{\left({k}−\mathrm{1}\right)!\left({n}−{k}+\mathrm{1}\right)!{k}}+\frac{{n}!{k}}{\left({k}−\mathrm{1}\right)!\left({n}−{k}+\mathrm{1}\right)!{k}}= \\ $$$$=\frac{{n}!\left({n}−{k}+\mathrm{1}\right)+{n}!{k}}{\left({k}−\mathrm{1}\right)!\left({n}−{k}+\mathrm{1}\right)!{k}}=\frac{{n}!\left({n}+\mathrm{1}\right)}{{k}!\left({n}+\mathrm{1}−{k}\right)!}=…

Question Number 71548 by Cmr 237 last updated on 17/Oct/19 $$\mathrm{let}\:\mathrm{f}:\boldsymbol{\mathrm{U}}\subset\mathbb{R}^{\mathrm{n}} \rightarrow\mathbb{R}^{\mathrm{p}} \:\mathrm{be}\:\mathrm{an}\:\mathrm{application} \\ $$$$\mathrm{where}\:\boldsymbol{\mathrm{U}}\:\mathrm{is}\:\mathrm{an}\:\mathrm{open}\:\mathrm{set} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\forall\mathrm{x}\in\boldsymbol{{U}},\exists\mathrm{h}\in\mathbb{R}^{\mathrm{n}} \mathrm{such}\:\mathrm{as} \\ $$$$\mathrm{x}+\mathrm{h}\in\boldsymbol{{U}} \\ $$ Answered by mind…