Question Number 5753 by Rasheed Soomro last updated on 26/May/16 $$\mathrm{If}\:\:\mathrm{0}<{r}<\mathrm{1},\:\mathrm{determine} \\ $$$$\mathrm{1}+\frac{{ar}}{{a}+{r}}+\frac{{ar}^{\mathrm{2}} }{{a}+\mathrm{2}{r}}+\frac{{ar}^{\mathrm{3}} }{{a}+\mathrm{3}{r}}….. \\ $$ Commented by FilupSmith last updated on 26/May/16 $${S}=\mathrm{1}+{T}…
Question Number 5752 by Rasheed Soomro last updated on 26/May/16 $$\mathrm{Determine} \\ $$$$\:\:\:\:\mathrm{1}+\frac{{ar}}{{a}+{r}}+\frac{{ar}^{\mathrm{2}} }{{a}+\mathrm{2}{r}}+\frac{{ar}^{\mathrm{3}} }{{a}+\mathrm{3}{r}}+….+\frac{{ar}^{{n}−\mathrm{1}} }{{a}+\left({n}−\mathrm{1}\right){r}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 71282 by mr W last updated on 13/Oct/19 $${solve} \\ $$$${x}\left({y}+{z}\right)=\mathrm{27} \\ $$$${y}\left({z}+{x}\right)=\mathrm{32} \\ $$$${z}\left({x}+{y}\right)=\mathrm{35} \\ $$ Answered by ajfour last updated on…
Question Number 5742 by Rasheed Soomro last updated on 26/May/16 $$\bullet{Determine}\:{S} \\ $$$${S}={a}^{\mathrm{2}} +{ar}\left({a}+{r}\right)+{ar}^{\mathrm{2}} \left({a}+\mathrm{2}{r}\right)+…+{ar}^{{n}−\mathrm{1}} \left\{{a}+\left({n}−\mathrm{1}\right)\:{r}\right\}. \\ $$$$ \\ $$ Answered by Yozzii last updated…
Question Number 5744 by Rasheed Soomro last updated on 26/May/16 $${If}\:\:\mathrm{0}<\:{r}<\mathrm{1},\:{is}\:{S}\:\:{convergent}\:{in}\:{the}\:{following}\:? \\ $$$${S}={a}^{\mathrm{2}} +{ar}\left({a}+{r}\right)+{ar}^{\mathrm{2}} \left({a}+\mathrm{2}{r}\right)+….. \\ $$$${Determine}\:{S}\:\:\:{if}\:\:{it}'{s}\:{convergent}. \\ $$ Commented by FilupSmith last updated on…
Question Number 5723 by Rasheed Soomro last updated on 25/May/16 $$\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}+….+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} }=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{n}} } \\ $$$$\mathrm{P}\:\:\:\mathrm{r}\:\:\:\mathrm{o}\:\:\:\mathrm{v}\:\:\:\mathrm{e}\:\mathrm{the}\:\mathrm{above}\:\mathrm{for}\:\mathrm{integral}\:\mathrm{n}\geqslant\mathrm{1}. \\ $$ Commented by FilupSmith last updated on 25/May/16 $$\boldsymbol{\mathrm{LHS}}={S}=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{8}}+…+\frac{\mathrm{1}}{\mathrm{2}^{{n}}…
Question Number 136793 by JulioCesar last updated on 26/Mar/21 Answered by Dwaipayan Shikari last updated on 26/Mar/21 $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{{x}^{{sin}\left({ax}\right)} }{{x}^{{tan}\left({bx}\right)} }\right)={y} \\ $$$$\Rightarrow\left({sin}\left({ax}\right)−{tan}\left({bx}\right)\right){log}\left({x}\right)={log}\left({y}\right) \\ $$$$\Rightarrow\left({acos}\left({ax}\right)−{bsec}^{\mathrm{2}}…
Question Number 5722 by Rasheed Soomro last updated on 25/May/16 $$\mathrm{Prove}\:\mathrm{by}\:\boldsymbol{\mathrm{mathematical}}\:\boldsymbol{\mathrm{induction}} \\ $$$$\mathrm{that}\:\mathrm{tbe}\:\mathrm{following}\:\mathrm{formula}\:\mathrm{is}\:\mathrm{correct} \\ $$$$\mathrm{for}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{n}: \\ $$$$\begin{pmatrix}{\mathrm{2}}\\{\mathrm{2}}\end{pmatrix}\:+\begin{pmatrix}{\mathrm{3}}\\{\mathrm{2}}\end{pmatrix}\:+\begin{pmatrix}{\mathrm{4}}\\{\mathrm{2}}\end{pmatrix}\:+…+\begin{pmatrix}{\mathrm{n}+\mathrm{1}}\\{\:\:\:\mathrm{2}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{n}+\mathrm{2}}\\{\:\:\:\mathrm{3}}\end{pmatrix} \\ $$ Commented by Yozzii last updated on…
Question Number 5712 by sanusihammed last updated on 24/May/16 $${Find}\:{the}\:{value}\:{of}\:{x}\: \\ $$$$ \\ $$$$\mathrm{2}^{{x}} \:=\:\mathrm{4}{x} \\ $$$$ \\ $$$${workings}\:{is}\:{needed}\:{please}. \\ $$ Commented by Yozzii last…
Question Number 71242 by TawaTawa last updated on 13/Oct/19 $$\mathrm{Given}:\:\:\:\frac{\mathrm{a}}{\mathrm{b}}\:+\:\frac{\mathrm{c}}{\mathrm{d}}\:\:=\:\:\frac{\mathrm{b}}{\mathrm{a}}\:+\:\frac{\mathrm{d}}{\mathrm{c}} \\ $$$$\mathrm{Show}\:\mathrm{that},\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{a}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }\:−\:\frac{\mathrm{c}^{\mathrm{2}} }{\mathrm{d}^{\mathrm{2}} }\:\:=\:\:\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\:−\:\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{c}^{\mathrm{2}} } \\ $$ Answered by MJS…