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Prove-n-0-a-n-b-n-c-n-n-c-c-a-b-c-a-c-b-Where-a-n-k-0-n-1-k-a-

Question Number 128321 by Dwaipayan Shikari last updated on 06/Jan/21 $${Prove} \\ $$$$\underset{{n}\geqslant\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({a}\right)_{{n}} \left({b}\right)_{{n}} }{\left({c}\right)_{{n}} {n}!}=\frac{\Gamma\left({c}\right)\Gamma\left({c}−{a}−{b}\right)}{\Gamma\left({c}−{a}\right)\Gamma\left({c}−{b}\right)} \\ $$$${Where}\:\left({a}\right)_{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}−\mathrm{1}} {\prod}}\left({k}+{a}\right) \\ $$…

An-element-X-has-RAM-of-88g-when-a-current-of-0-5A-was-passed-through-fused-chloride-of-X-for-32minutes-and-10sec-0-44g-of-X-was-deposited-at-the-cathode-a-number-of-faraday-b-write-formular-of-X

Question Number 62750 by peter frank last updated on 24/Jun/19 $${An}\:{element}\:{X}\:{has}\:{RAM} \\ $$$${of}\:\mathrm{88}{g}.{when}\:{a}\:{current} \\ $$$${of}\:\mathrm{0}.\mathrm{5}{A}\:{was}\:{passed}\:{through} \\ $$$${fused}\:{chloride}\:{of}\:{X}\:{for} \\ $$$$\mathrm{32}{minutes}\:{and}\:\mathrm{10}{sec}. \\ $$$$\mathrm{0}.\mathrm{44}{g}\:{of}\:{X}\:{was}\:{deposited} \\ $$$${at}\:{the}\:{cathode} \\ $$$$\left({a}\right){number}\:{of}\:{faraday}?…

1-1-2-1-1-2-2-1-3-2-1-3-2-2-2-1-4-3-1-3-5-2-3-2-4-pi-Prove-the-above-Relation-

Question Number 128256 by Dwaipayan Shikari last updated on 05/Jan/21 $$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}!\mathrm{1}!}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{3}!\mathrm{2}!}\left(\frac{\mathrm{1}.\mathrm{3}}{\mathrm{2}^{\mathrm{2}} }\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}!\mathrm{3}!}\left(\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}}{\mathrm{2}^{\mathrm{3}} }\right)^{\mathrm{2}} +….=\frac{\mathrm{4}}{\pi} \\ $$$${Prove}\:{the}\:{above}\:{Relation} \\ $$ Commented by Dwaipayan Shikari last…

1-6-p-logp-p-2-1-log1-pi-2-log2-4pi-2-log3-9pi-2-p-prime-

Question Number 128236 by Dwaipayan Shikari last updated on 05/Jan/21 $$\frac{\mathrm{1}}{\mathrm{6}}\underset{{p}} {\overset{\infty} {\sum}}\frac{{logp}}{{p}^{\mathrm{2}} −\mathrm{1}}=\frac{{log}\mathrm{1}}{\pi^{\mathrm{2}} }+\frac{{log}\mathrm{2}}{\mathrm{4}\pi^{\mathrm{2}} }+\frac{{log}\mathrm{3}}{\mathrm{9}\pi^{\mathrm{2}} }+…\:\:\:\left({p}={prime}\right) \\ $$ Commented by Dwaipayan Shikari last updated…

4-99-7-999-11-999999-

Question Number 128126 by Agnibhoo last updated on 04/Jan/21 $$\:\frac{\mathrm{4}}{\mathrm{99}}\:+\:\frac{\mathrm{7}}{\mathrm{999}}\:+\:\frac{\mathrm{11}}{\mathrm{999999}}\:=\:? \\ $$ Answered by Geovanek last updated on 04/Jan/21 $$\frac{\mathrm{4}}{\mathrm{99}}\:+\:\frac{\mathrm{7}}{\mathrm{99}}\:+\:\frac{\mathrm{11}}{\mathrm{999999}}\:=\:{X} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{see}\:\mathrm{that} \\ $$$$\frac{\mathrm{999999}}{\mathrm{99}}\:=\:\mathrm{10101}\:\:\:\boldsymbol{\mathrm{AND}} \\…

1-1-16-5-2-16-2-2-5-2-9-2-16-3-3-5-2-9-2-13-2-16-4-4-pi-2-3-4-F-1-1-4-1-4-1-1-Prove-The-above-relation-Where-F-1-n-0-n-

Question Number 128122 by Dwaipayan Shikari last updated on 04/Jan/21 $$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{16}}+\frac{\mathrm{5}^{\mathrm{2}} }{\mathrm{16}^{\mathrm{2}} .\mathrm{2}!}+\frac{\mathrm{5}^{\mathrm{2}} .\mathrm{9}^{\mathrm{2}} }{\mathrm{16}^{\mathrm{3}} .\mathrm{3}!}+\frac{\mathrm{5}^{\mathrm{2}} .\mathrm{9}^{\mathrm{2}} .\mathrm{13}^{\mathrm{2}} }{\mathrm{16}^{\mathrm{4}} .\mathrm{4}!}+…=\frac{\sqrt{\pi}}{\Gamma^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)}={F}_{\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{4}},\mathrm{1};\mathrm{1}\right) \\ $$$${Prove}\:{The}\:{above}\:{relation} \\…

1-2-3-4-100-

Question Number 128112 by AgnibhoMukhopadhyay last updated on 04/Jan/21 $$\:\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:\mathrm{4}\:+\:…..\:+\:\mathrm{100}\:=\:? \\ $$ Answered by Olaf last updated on 04/Jan/21 $$\mathrm{A}\:\left({n}+\mathrm{1}\right)×\left({n}+\mathrm{1}\right)\:\mathrm{squared}\:\mathrm{chess}\:\mathrm{board} \\ $$$$\mathrm{contains}\:\left({n}+\mathrm{1}\right)^{\mathrm{2}} \:\mathrm{squares}. \\ $$$$…