Menu Close

Author: Tinku Tara

Question-132205

Question Number 132205 by Arijit last updated on 12/Feb/21 Commented by Arijit last updated on 12/Feb/21 $$\boldsymbol{\mathrm{Please}}\:\boldsymbol{\mathrm{Help}}\:\boldsymbol{\mathrm{me}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{this}}….. \\ $$ Answered by mathmax by abdo last…

Question-66670

Question Number 66670 by naka3546 last updated on 18/Aug/19 Commented by mathmax by abdo last updated on 18/Aug/19 $${let}\:{S}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{3}^{{n}} }{\mathrm{5}^{{n}} \left({n}^{\mathrm{2}} \:+\mathrm{3}{n}+\mathrm{2}\right)}\:\Rightarrow{S}\:=\sum_{{n}=\mathrm{0}} ^{\infty}…

Find-the-infinite-product-of-3-2-5-4-17-16-257-256-65537-65536-

Question Number 1134 by 314159 last updated on 28/Jun/15 $${Find}\:{the}\:{infinite}\:{product}\:{of}\: \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}×\frac{\mathrm{5}}{\mathrm{4}}×\frac{\mathrm{17}}{\mathrm{16}}×\frac{\mathrm{257}}{\mathrm{256}}×\frac{\mathrm{65537}}{\mathrm{65536}}×… \\ $$ Answered by prakash jain last updated on 29/Jun/15 $$\frac{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}^{\mathrm{2}}…

Let-f-0-1-R-be-a-differentiable-function-Prove-that-there-exists-a-c-0-1-such-that-4-pi-f-1-f-0-1-c-2-f-c-

Question Number 1133 by 112358 last updated on 29/Jun/15 $${Let}\:{f}\::\:\left[\:\mathrm{0}\:,\:\mathrm{1}\:\right]\:\rightarrow\:\mathbb{R}\:\:{be}\:{a}\: \\ $$$${differentiable}\:{function}.\:{Prove} \\ $$$${that}\:{there}\:{exists}\:{a}\:{c}\:\in\:\left[\mathrm{0},\mathrm{1}\right]\:{such} \\ $$$${that}\: \\ $$$$\frac{\mathrm{4}}{\pi}\left[{f}\left(\mathrm{1}\right)−{f}\left(\mathrm{0}\right)\right]=\left(\mathrm{1}+{c}^{\mathrm{2}} \right){f}^{\:} '\left({c}\right).\: \\ $$ Commented by 123456…

let-S-1-2-3-4-5-if-A-B-C-is-such-that-A-B-C-A-B-A-C-how-many-ways-can-be-choose-A-B-and-C-

Question Number 1132 by 123456 last updated on 24/Jun/15 $$\mathrm{let}\:\mathrm{S}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5}\right\},\:\mathrm{if}\:\mathrm{A},\mathrm{B},\mathrm{C}\:\mathrm{is}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{A}\cap\mathrm{B}\cap\mathrm{C}=\varnothing \\ $$$$\mathrm{A}\cap\mathrm{B}\neq\varnothing \\ $$$$\mathrm{A}\cap\mathrm{C}\neq\varnothing \\ $$$$\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{be}\:\mathrm{choose}\:\mathrm{A},\mathrm{B}\:\mathrm{and} \\ $$$$\mathrm{C} \\ $$ Commented by Faaiz…