Question Number 56886 by gunawan last updated on 25/Mar/19 $$\underset{{r}=\mathrm{0}} {\overset{{n}} {\sum}}\:^{{n}} {C}_{{r}} \frac{\mathrm{1}+{r}\:\mathrm{log}_{{e}} \mathrm{10}}{\left(\mathrm{1}+\mathrm{log}_{{e}} \mathrm{10}^{{n}} \right)^{{r}} }\:\:\mathrm{equals} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated…
Question Number 56424 by gunawan last updated on 16/Mar/19 $$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{first}\:\mathrm{10}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series} \\ $$$$\left({x}+\:\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} \:+\:\left({x}^{\mathrm{2}} +\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \:+\:\left({x}^{\mathrm{3}} +\:\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\mathrm{2}} +…\:\mathrm{is} \\ $$ Answered by mr W…
Question Number 56423 by gunawan last updated on 16/Mar/19 $$\mathrm{The}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{infinity}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{15}}\:+\:\frac{\mathrm{1}}{\mathrm{35}}\:+\:….\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{2}}. \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 16/Mar/19 $${T}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{3}−\mathrm{1}}{\mathrm{3}×\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}}\right) \\ $$$${T}_{\mathrm{2}}…
Question Number 56422 by gunawan last updated on 16/Mar/19 $$\mathrm{If}\:\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations} \\ $$$${ax}\:+\:{by}\:+\:\left({a}\lambda+{b}\right){z}\:=\:\mathrm{0} \\ $$$${bx}\:+\:{cy}\:+\:\left({b}\lambda+{c}\right){z}\:=\:\mathrm{0} \\ $$$$\left({a}\lambda\:+\:{b}\right){x}\:+\:\left({b}\lambda+{c}\right){y}\:=\:\mathrm{0} \\ $$$$\mathrm{has}\:\mathrm{a}\:\mathrm{non}−\mathrm{trivial}\:\mathrm{solution},\:\mathrm{then} \\ $$ Answered by MJS last updated…
Question Number 56421 by gunawan last updated on 16/Mar/19 $$\mathrm{Let}\:{a}_{\mathrm{1}} ,\:{a}_{\mathrm{2}} \:,\:…,\:{a}_{\mathrm{10}} \:\mathrm{be}\:\mathrm{in}\:\mathrm{AP}\:\mathrm{and}\:{h}_{\mathrm{1}} ,\:{h}_{\mathrm{2}} ,…,\:{h}_{\mathrm{10}} \\ $$$$\mathrm{be}\:\mathrm{in}\:\mathrm{HP}.\:\mathrm{If}\:\:{a}_{\mathrm{1}} =\:{h}_{\mathrm{1}} =\mathrm{2}\:\:\mathrm{and}\:\:{a}_{\mathrm{10}} =\:{h}_{\mathrm{10}} =\mathrm{3}, \\ $$$$\mathrm{then}\:{a}_{\mathrm{4}} {h}_{\mathrm{7}} \:\:\mathrm{is}…
Question Number 56420 by gunawan last updated on 16/Mar/19 $$\mathrm{Let}\:\:{a},\:{b}\:\mathrm{and}\:{c}\:\mathrm{form}\:\mathrm{a}\:\mathrm{GP}\:\mathrm{of}\:\mathrm{common}\:\mathrm{ratio}\:{r}, \\ $$$$\mathrm{with}\:\:\mathrm{0}<\:{r}<\mathrm{1}.\:\mathrm{If}\:\:{a},\:\mathrm{2}{b}\:\mathrm{and}\:\mathrm{3}{c}\:\mathrm{form}\:\mathrm{an}\:\mathrm{AP}, \\ $$$$\mathrm{then}\:{r}\:\mathrm{equals} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 16/Mar/19 $${a}={a} \\…
Question Number 56419 by gunawan last updated on 16/Mar/19 $$\mathrm{Let}\:{a},\:{b},\:{c}\:\mathrm{be}\:\mathrm{in}\:\mathrm{AP}\:\mathrm{and}\:\mid{a}\mid<\mathrm{1},\:\mid{b}\mid<\mathrm{1},\:\mid{c}\mid<\mathrm{1}.\:\mathrm{If} \\ $$$${x}\:\:=\:\:\mathrm{1}+{a}+{a}^{\mathrm{2}} +…\:\mathrm{to}\:\infty\: \\ $$$${y}\:\:=\:\:\mathrm{1}+{b}+{b}^{\mathrm{2}} +…\:\mathrm{to}\:\infty\: \\ $$$${z}\:\:=\:\:\mathrm{1}+{c}+{c}^{\mathrm{2}} +…\:\mathrm{to}\:\infty\:\: \\ $$$$\mathrm{then}\:{x},\:{y},\:{z}\:\mathrm{are}\:\mathrm{in} \\ $$ Answered by…
Question Number 56418 by gunawan last updated on 16/Mar/19 $$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\:{a}^{\mathrm{log}_{{b}} {x}} \mathrm{where}\:{a}=\mathrm{0}.\mathrm{2},\:{b}=\sqrt{\mathrm{5},} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{8}}\:+\:\frac{\mathrm{1}}{\mathrm{16}}\:+\:…\:\mathrm{to}\:\infty\:\mathrm{is} \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 16/Mar/19 $${x}=\frac{\frac{\mathrm{1}}{\mathrm{4}}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{2}}=\mathrm{0}.\mathrm{5} \\…
Question Number 56417 by gunawan last updated on 16/Mar/19 $$\mathrm{The}\:{n}\mathrm{th}\:\mathrm{term}\:\mathrm{of}\:\mathrm{the}\:\mathrm{series}\: \\ $$$$\mathrm{4},\:\mathrm{14},\:\mathrm{30},\:\mathrm{52},\:\mathrm{80},\:\mathrm{114},\:…\:\:\mathrm{is} \\ $$ Answered by MJS last updated on 16/Mar/19 $$\mathrm{4}\:\:\:\:\:\mathrm{14}\:\:\:\:\:\mathrm{30}\:\:\:\:\:\mathrm{52}\:\:\:\:\:\mathrm{80}\:\:\:\:\:\mathrm{114} \\ $$$$\:\:\:\mathrm{10}\:\:\:\:\:\mathrm{16}\:\:\:\:\:\mathrm{22}\:\:\:\:\:\mathrm{28}\:\:\:\:\:\mathrm{34} \\…
Question Number 56416 by gunawan last updated on 16/Mar/19 $$\underset{{n}−\mathrm{digits}} {\left(\mathrm{666}\:….\:\mathrm{6}\right)^{\mathrm{2}} }\:+\:\underset{{n}−\mathrm{digits}} {\left(\mathrm{888}\:….\mathrm{8}\right)}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to} \\ $$ Commented by mr W last updated on 16/Mar/19 $$=\underset{\mathrm{2}{n}\:{digits}} {\left(\mathrm{444}…\mathrm{4}\right)}…