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Author: Tinku Tara

S-n-1-1-n-1-H-n-n-2-note-H-n-1-1-2-1-3-1-n-

Question Number 222478 by mnjuly1970 last updated on 28/Jun/25 $$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \frac{{H}_{{n}} }{{n}^{\mathrm{2}} }\:=\:? \\ $$$$\:{note}:\:\:\:{H}_{{n}} =\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}}\:+…+\frac{\mathrm{1}}{{n}}\: \\ $$ Answered by MrGaster…

Is-the-statement-correct-in-const-x-a-bx-n-0-x-a-2b-n-1-n-gt-0-N-b-a-b-a-G-n-1-n-0-

Question Number 222511 by hu last updated on 28/Jun/25 $${Is}\:{the}\:{statement}\:{correct}? \\ $$$${in}\:{const}.“\:{x}+{a}−{bx}^{{n}} =\left\{\mathrm{0}\right\}\:'' \\ $$$${x}=\begin{cases}{−{a}\pm\frac{\mathrm{2}{b}}{{n}+\mathrm{1}}\:,\:{n}>\mathrm{0}}\\{{N}_{{b}} ^{\:{a}} \:\int_{{b}} ^{\:{a}} {G}\left({n}−\mathrm{1}\right),\:{n}\leqslant\mathrm{0}}\end{cases} \\ $$ Terms of Service Privacy…

This-is-VERY-HARD-x-y-0-l-y-1-x-N-y-v-for-x-1-v-for-x-x-0-c-7-x-y-

Question Number 222501 by hu last updated on 28/Jun/25 $${This}\:{is}\:{VERY}\:{HARD} \\ $$$$\begin{cases}{\begin{cases}{{x}+{y}=\mathrm{0}}\\{{l}\left({y}\right)=\mathrm{1}}\end{cases}}\\{\begin{cases}{{x}\in\mathbb{N}}\\{−{y}=\begin{cases}{{v\%},\:\:{for}\:{x}\circlearrowleft\gamma\left(\mathrm{1}\right)}\\{−{v\%},\:\:{for}\:{x}\looparrowright\theta\left(\oint_{−{x}} ^{\:\mathrm{0}} \frac{{c}}{\mathrm{7}}\right)}\end{cases}}\end{cases}}\end{cases} \\ $$$${x}=?,\:{y}=? \\ $$ Answered by wewji12 last updated on 28/Jun/25…

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Question Number 222462 by fantastic last updated on 27/Jun/25 $${i}^{{i}} =?? \\ $$ Answered by vnm last updated on 27/Jun/25 $$=\left({e}^{{i}\left(\frac{\pi}{\mathrm{2}}+\mathrm{2}{n}\pi\right)} \right)^{{i}} ={e}^{\pi\left(−\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{2}{n}\right)} ,\:{n}\in\mathbb{Z} \\…

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Question Number 222441 by fantastic last updated on 27/Jun/25 $$\mathrm{0}^{{i}} \\ $$ Answered by wewji12 last updated on 27/Jun/25 $$\mathrm{0}\:\mathrm{because}\:{f}\left({z}\right)=\mathrm{0}^{{z}} \:\mathrm{for}\:\mathrm{all}\:{z}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\} \\ $$$${f}\left({z}\right)=\mathrm{0}\:\mathrm{const}. \\ $$$${f}\left({z}\right)=\begin{cases}{\:\mathrm{1}\:,\:{z}=\mathrm{0}}\\{\:\mathrm{0}\:,\:{z}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\}}\end{cases}…