Question Number 10169 by Joel575 last updated on 28/Jan/17 $$\frac{\mathrm{2013}}{\mathrm{1}}\:+\:\frac{\mathrm{2013}}{\mathrm{1}+\mathrm{2}}\:+\:\frac{\mathrm{2013}}{\mathrm{1}+\mathrm{2}+\mathrm{3}}\:+\:…\:+\:\frac{\mathrm{2013}}{\mathrm{1}+\mathrm{2}+\mathrm{3}+…+\mathrm{2012}}\:=\:? \\ $$ Answered by prakash jain last updated on 29/Jan/17 $$\underset{{i}=\mathrm{1}} {\overset{\mathrm{2012}} {\sum}}\:\frac{\mathrm{2013}}{\underset{{j}=\mathrm{1}} {\overset{{i}} {\sum}}{j}}\:=\:\:\mathrm{2013}\underset{{i}=\mathrm{1}}…
Question Number 10154 by Tawakalitu ayo mi last updated on 27/Jan/17 $$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{first}\:\mathrm{term}\:\mathrm{of}\:\mathrm{sequence}\:\mathrm{is}\: \\ $$$$\mathrm{given}\:\mathrm{by}\:\:\mathrm{S}_{\mathrm{n}} \:=\:\mathrm{5n}^{\mathrm{2}} \:−\:\mathrm{2n}.\:\mathrm{A}\:\mathrm{sequence}\:\: \\ $$$$\mathrm{U}_{\mathrm{1}} ,\:\mathrm{U}_{\mathrm{2}} ,\:\mathrm{U}_{\mathrm{3}} \:….\:\mathrm{is}\:\mathrm{defined}\:\mathrm{by}\:\mathrm{U}_{\mathrm{t}} \:=\:\mathrm{S}_{\mathrm{t}} \:−\:\mathrm{S}_{\mathrm{t}\:−\:\mathrm{1}} . \\…
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Question Number 75641 by aliesam last updated on 14/Dec/19 Commented by vishalbhardwaj last updated on 15/Dec/19 $$\mathrm{Hypergeometric}\:\mathrm{Function}\::\: \\ $$$$\mathrm{2F1}\:\left({a},{b};{c};{z}\right)\:=\:\mathrm{1}+\frac{{ab}}{\mathrm{1}!\:{c}}\:{z}+\frac{{a}\left({a}+\mathrm{1}\right){b}\left({b}+\mathrm{1}\right)}{\mathrm{2}!\:{c}\left({c}+\mathrm{1}\right)}\:{z}^{\mathrm{2}} +\frac{{a}\left({a}+\mathrm{1}\right)\left({a}+\mathrm{2}\right){b}\left({b}+\mathrm{1}\right)\left({b}+\mathrm{2}\right)}{\mathrm{3}!\:{c}\left({c}+\mathrm{1}\right)\left({c}+\mathrm{2}\right)}\:{z}^{\mathrm{3}} +\:.\:.\:. \\ $$ Answered by…
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Question Number 141104 by Opredador last updated on 15/May/21 Answered by hknkrc46 last updated on 15/May/21 $$\left.\begin{matrix}{\sqrt{\mathrm{7}^{\sqrt{\mathrm{63}}} }\:=\:\sqrt{\mathrm{7}^{\mathrm{3}\sqrt{\mathrm{7}}} }\:=\:\mathrm{7}^{\sqrt{\mathrm{7}}} \sqrt{\mathrm{7}^{\sqrt{\mathrm{7}}} }}\\{\mathrm{7}^{\sqrt{\mathrm{7}}} \:=\:\boldsymbol{{u}}}\end{matrix}\right\}\:\frac{\boldsymbol{{u}}\sqrt{\boldsymbol{{u}}}\:−\:\sqrt{\boldsymbol{{u}}}}{\boldsymbol{{u}}\:−\:\mathrm{1}} \\ $$$$=\:\frac{\sqrt{\boldsymbol{{u}}}\left(\boldsymbol{{u}}\:−\:\mathrm{1}\right)}{\boldsymbol{{u}}\:−\:\mathrm{1}}\:=\:\sqrt{\boldsymbol{{u}}}\:=\:\sqrt{\mathrm{7}^{\sqrt{\mathrm{7}}} }…
Question Number 75440 by aliesam last updated on 11/Dec/19 Commented by MJS last updated on 11/Dec/19 $$\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{good}\:\mathrm{at}\:\mathrm{this}\:\mathrm{but}\:\mathrm{it}\:\mathrm{should}\:\mathrm{be}\:\mathrm{true} \\ $$$$\mathrm{for}\:\mathrm{any}\:{x}\:\mathrm{at}\:\mathrm{least}\:{x}\geqslant\mathrm{1} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{k}} \mathrm{e}^{−\left(\mathrm{2}{k}−\mathrm{1}\right){x}} }{\left(\mathrm{2}{k}−\mathrm{1}\right)\left(\mathrm{1}−\mathrm{e}^{−\mathrm{2}{x}}…
Question Number 75421 by peter frank last updated on 10/Dec/19 $${If}\:{p}\:{is}\:{a}\:{point}\:{in}\:{the}\:{base} \\ $$$${AB}\:{of}\:\:{a}\:\:{triangle}\:\:{ABC} \\ $$$${such}\:{that}\:{AP}\:\::{PB}={P}:{Q} \\ $$$${prove}\:{that} \\ $$$$\left({p}+{q}\right)\mathrm{cot}\:\theta={q}\mathrm{cot}\:{A}−{p}\mathrm{cot}\:{B} \\ $$ Commented by som(math1967) last…