Question Number 50367 by prof Abdo imad last updated on 16/Dec/18 $${calculate}\:\sum_{{k}={p}} ^{\mathrm{2}{p}} \:\:\frac{{C}_{{k}} ^{{p}} }{\mathrm{2}^{{k}} } \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 50366 by prof Abdo imad last updated on 16/Dec/18 $${let}\:\:{A}\:=\begin{pmatrix}{\mathrm{0}\:\:\:\:\:{m}\:\:\:\:\:\:{m}^{\mathrm{2}} }\\{\frac{\mathrm{1}}{{m}}\:\:\:\:\:\mathrm{0}\:\:\:\:\:\:\:{m}}\end{pmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\frac{\mathrm{1}}{{m}^{\mathrm{2}} }\:\:\:\:\frac{\mathrm{1}}{{m}}\:\:\:\:\:\:\mathrm{0}\:\:\:\right) \\ $$$${A}\:\in\:{M}_{\mathrm{3}} \left({R}\right)\:\:{and}\:{m}\:{not}\:\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{relation}\:{betwen}\:{I}_{\mathrm{3}} ,\:{A}\:{and}\:{A}^{\mathrm{2}} \\ $$$$\left.\mathrm{2}\right)\:{is}\:{A}\:{inversible}\:\:\:.{determine}\:{A}^{−\mathrm{1}} \:{in}\:{case}\:{of}\:{exist}…
Question Number 181438 by Agnibhoo98 last updated on 26/Nov/22 $$\mathrm{If}\:{a}\:+\:\frac{\mathrm{1}}{{b}}\:=\:{b}\:+\:\frac{\mathrm{1}}{{c}}\:=\:{c}\:+\:\frac{\mathrm{1}}{{a}}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${abc}\:=\:\pm\mathrm{1}.\:\:\:{a}\:\neq\:{b}\:\neq\:{c} \\ $$ Commented by mr W last updated on 26/Nov/22 $${it}'{s}\:{not}\:{true}.\:{e}.{g}.\:{you}\:{can}\:{take}\: \\ $$$${a}={b}={c}=\mathrm{2},\:{and}\:{get}\:{abc}=\mathrm{8}.…
Question Number 50365 by prof Abdo imad last updated on 16/Dec/18 $${let}\:{J}\:=\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}}\end{bmatrix} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{bmatrix}{\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}}\\{}\end{bmatrix} \\ $$$${element}\:{of}\:{M}_{\mathrm{3}} \left({R}\right) \\ $$$${find}\:{J}^{{n}} \\ $$$$ \\ $$ Terms of…
Question Number 50364 by prof Abdo imad last updated on 16/Dec/18 $${let}\:{A}\:=\begin{pmatrix}{\mathrm{2}\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\mathrm{2}}\end{pmatrix} \\ $$$${calculate}\:{A}^{{n}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 50363 by prof Abdo imad last updated on 16/Dec/18 $${let}\:{p}\:\in{K}_{{n}} \left[{x}\right]\:{snd}\:{A}\:{and}\:{H}\:{two}\:{rlements}\:{of}\:{K}\left[{x}\right] \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{p}\left({A}\left({x}\right)+{H}\left({x}\right)\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \frac{{p}^{\left({k}\right)} \left({A}\left({x}\right)\right)}{{k}!}.\left({H}\left({x}\right)\right)^{{k}} \\ $$$$\left.\mathrm{2}\right){find}\:{the}\:{condition}\:{that}\:{p}\left({A}\left({x}\right)+{H}\left({x}\right)\right){is} \\ $$$${divided}\:{by}\:{H}\left({x}\right)\neq\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:{if}\:{p}\left({x}\right)\neq{c}\:{prove}\:\:{that}\:{p}\left({p}\left({x}\right)\right)−{x}\:{is}\:{divided} \\…
Question Number 50362 by prof Abdo imad last updated on 16/Dec/18 $${calculate}\:{S}_{\mathrm{1}} =\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \\ $$$${S}_{\mathrm{2}} =\sum_{{k}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\:{C}_{{n}} ^{\mathrm{2}{k}} \\ $$$${S}_{\mathrm{3}} =\:\sum_{{k}=\mathrm{0}}…
Question Number 115897 by Joel574 last updated on 29/Sep/20 $$\mathrm{Given} \\ $$$${a}_{{n}} \:=\:\sqrt{\mathrm{1}\:+\:\left(\mathrm{1}\:−\:\frac{\mathrm{1}}{{n}}\right)^{\mathrm{2}} }\:+\:\sqrt{\mathrm{1}\:+\:\left(\mathrm{1}\:+\:\frac{\mathrm{1}}{{n}}\right)^{\mathrm{2}} } \\ $$$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{2015}} {\sum}}\left(\frac{\mathrm{4}}{{a}_{{n}} }\right)\:\mathrm{is}\:… \\ $$ Commented by Joel574…
Question Number 50359 by prof Abdo imad last updated on 16/Dec/18 $$\left.\mathrm{1}\right)\:{calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{{k}} \:{C}_{{n}} ^{{k}} \:\:\frac{\mathrm{1}}{{k}+\mathrm{1}} \\ $$$$\left.\mathrm{2}\right){calculate}\:{S}_{{n}} \left({p}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \:\:\:\frac{{C}_{{n}} ^{{k}} }{{p}+{k}+\mathrm{1}}…
Question Number 50355 by prof Abdo imad last updated on 16/Dec/18 $${let}\:\left(\alpha_{{k}} \right)\:\:\left({k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right]\:{the}\:{n}^{{eme}} \:{roots}\:{of}\:\mathrm{1}\right. \\ $$$${calculate}\:{p}\left({x},{y}\right)=\left({x}+\alpha_{\mathrm{0}} {y}\right)\left({x}+\alpha_{\mathrm{1}} {y}\right)….\left({x}+\alpha_{{n}−\mathrm{1}} {y}\right) \\ $$ Terms of Service Privacy…