Menu Close

prove-that-2x-4-2-3-4-3x-5-3-4-5-4x-6-4-5-6-100x-102-100-101-102-103-102-




Question Number 191790 by mathlove last updated on 30/Apr/23
prove that  ((2x−4)/(2∙3∙4))+((3x−5)/(3∙4∙5))+((4x−6)/(4∙5∙6))+.....+((100x−102)/(100∙101∙102))=((103)/(102))
$${prove}\:{that} \\ $$$$\frac{\mathrm{2}{x}−\mathrm{4}}{\mathrm{2}\centerdot\mathrm{3}\centerdot\mathrm{4}}+\frac{\mathrm{3}{x}−\mathrm{5}}{\mathrm{3}\centerdot\mathrm{4}\centerdot\mathrm{5}}+\frac{\mathrm{4}{x}−\mathrm{6}}{\mathrm{4}\centerdot\mathrm{5}\centerdot\mathrm{6}}+…..+\frac{\mathrm{100}{x}−\mathrm{102}}{\mathrm{100}\centerdot\mathrm{101}\centerdot\mathrm{102}}=\frac{\mathrm{103}}{\mathrm{102}} \\ $$$$ \\ $$
Commented by BaliramKumar last updated on 01/May/23
if x=1 then all term (−ve) but ((103)/(102)) is (+ve)
$$\mathrm{if}\:\mathrm{x}=\mathrm{1}\:\mathrm{then}\:\mathrm{all}\:\mathrm{term}\:\left(−\mathrm{ve}\right)\:\mathrm{but}\:\frac{\mathrm{103}}{\mathrm{102}}\:\mathrm{is}\:\left(+\mathrm{ve}\right)\: \\ $$
Commented by mathlove last updated on 01/May/23
waht is sulotion
$${waht}\:{is}\:{sulotion} \\ $$
Commented by mathlove last updated on 01/May/23
Q type is right
$${Q}\:{type}\:{is}\:{right} \\ $$
Commented by mehdee42 last updated on 01/May/23
please check the correctness of the givrn statment again
$${please}\:{check}\:{the}\:{correctness}\:{of}\:{the}\:{givrn}\:{statment}\:{again} \\ $$
Commented by mr W last updated on 01/May/23
it is shown that the question is wrong!  just put x=0 or x=1!  you can not prove something which  is not true!
$${it}\:{is}\:{shown}\:{that}\:{the}\:{question}\:{is}\:{wrong}! \\ $$$${just}\:{put}\:{x}=\mathrm{0}\:{or}\:{x}=\mathrm{1}! \\ $$$${you}\:{can}\:{not}\:{prove}\:{something}\:{which} \\ $$$${is}\:{not}\:{true}! \\ $$
Commented by BaliramKumar last updated on 01/May/23
you mean find the value of  x ?
$$\mathrm{you}\:\mathrm{mean}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:{x}\:? \\ $$
Commented by mathlove last updated on 02/May/23
yes
$${yes} \\ $$
Answered by BaliramKumar last updated on 01/May/23
((2x)/(2×3×4)) − (4/(2×3×4)) + ((3x)/(3×4×5)) − (5/(3×4×5)) + ...........+ ((100x)/(100×101×102)) − ((102)/(100×101×102)) = ((103)/(102))              (x/(3×4)) − (1/(2×3)) + (x/(4×5)) − (1/(3×4)) + ...........+ (x/(101×102)) − (1/(100×101)) = ((103)/(102))  (x/(3×4)) − (1/(2×3)) + (x/(4×5)) − (1/(3×4)) + ...........+ (x/(101×102)) − (1/(100×101)) = ((103)/(102))                       x[(1/3) − (1/(102))] −[(1/2) − (1/(101))] = ((103)/(102))  x[ ((34−1)/(102))] −[((101−2)/(2×101))] = ((103)/(102))  x[ ((33)/(102))] −[((99)/(202))] = ((103)/(102))  x[ ((11)/(34))]  = ((103)/(102)) + ((99)/(202))  ((11x)/(34)) = ((7726)/(5151))  x = ((15452)/(3333))
$$\frac{\mathrm{2}{x}}{\mathrm{2}×\mathrm{3}×\mathrm{4}}\:−\:\frac{\mathrm{4}}{\mathrm{2}×\mathrm{3}×\mathrm{4}}\:+\:\frac{\mathrm{3}{x}}{\mathrm{3}×\mathrm{4}×\mathrm{5}}\:−\:\frac{\mathrm{5}}{\mathrm{3}×\mathrm{4}×\mathrm{5}}\:+\:………..+\:\frac{\mathrm{100}{x}}{\mathrm{100}×\mathrm{101}×\mathrm{102}}\:−\:\frac{\mathrm{102}}{\mathrm{100}×\mathrm{101}×\mathrm{102}}\:=\:\frac{\mathrm{103}}{\mathrm{102}}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\frac{{x}}{\mathrm{3}×\mathrm{4}}\:−\:\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}}\:+\:\frac{{x}}{\mathrm{4}×\mathrm{5}}\:−\:\frac{\mathrm{1}}{\mathrm{3}×\mathrm{4}}\:+\:………..+\:\frac{{x}}{\mathrm{101}×\mathrm{102}}\:−\:\frac{\mathrm{1}}{\mathrm{100}×\mathrm{101}}\:=\:\frac{\mathrm{103}}{\mathrm{102}} \\ $$$$\frac{{x}}{\mathrm{3}×\mathrm{4}}\:−\:\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}}\:+\:\frac{{x}}{\mathrm{4}×\mathrm{5}}\:−\:\frac{\mathrm{1}}{\mathrm{3}×\mathrm{4}}\:+\:………..+\:\frac{{x}}{\mathrm{101}×\mathrm{102}}\:−\:\frac{\mathrm{1}}{\mathrm{100}×\mathrm{101}}\:=\:\frac{\mathrm{103}}{\mathrm{102}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$${x}\left[\frac{\mathrm{1}}{\mathrm{3}}\:−\:\frac{\mathrm{1}}{\mathrm{102}}\right]\:−\left[\frac{\mathrm{1}}{\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{101}}\right]\:=\:\frac{\mathrm{103}}{\mathrm{102}} \\ $$$${x}\left[\:\frac{\mathrm{34}−\mathrm{1}}{\mathrm{102}}\right]\:−\left[\frac{\mathrm{101}−\mathrm{2}}{\mathrm{2}×\mathrm{101}}\right]\:=\:\frac{\mathrm{103}}{\mathrm{102}} \\ $$$${x}\left[\:\frac{\mathrm{33}}{\mathrm{102}}\right]\:−\left[\frac{\mathrm{99}}{\mathrm{202}}\right]\:=\:\frac{\mathrm{103}}{\mathrm{102}} \\ $$$${x}\left[\:\frac{\mathrm{11}}{\mathrm{34}}\right]\:\:=\:\frac{\mathrm{103}}{\mathrm{102}}\:+\:\frac{\mathrm{99}}{\mathrm{202}} \\ $$$$\frac{\mathrm{11}{x}}{\mathrm{34}}\:=\:\frac{\mathrm{7726}}{\mathrm{5151}} \\ $$$${x}\:=\:\frac{\mathrm{15452}}{\mathrm{3333}} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *