prove that ((2x−4)/(2∙3∙4))+((3x−5)/(3∙4∙5))+((4x−6)/(4∙5∙6))+.....+((100x−102)/(100∙101∙102))=((103)/(102))


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Question Number 191790 by mathlove last updated on 30/Apr/23

$p r o v e t h a t$ $\frac{2 x - 4}{2 \cdot 3 \cdot 4} + \frac{3 x - 5}{3 \cdot 4 \cdot 5} + \frac{4 x - 6}{4 \cdot 5 \cdot 6} + . . . . . + \frac{100 x - 102}{100 \cdot 101 \cdot 102} = \frac{103}{102}$
Commented by BaliramKumar last updated on 01/May/23

$if x = 1 then all term (- ve) but \frac{103}{102} is (+ ve)$
Commented by mathlove last updated on 01/May/23

$w a h t i s s u l o t i o n$
Commented by mathlove last updated on 01/May/23

$Q t y p e i s r i g h t$
Commented by mehdee42 last updated on 01/May/23

$p l e a s e c h e c k t h e c o r r e c t n e s s o f t h e g i v r n s t a t m e n t a g a i n$
Commented by mr W last updated on 01/May/23

$i t i s s h o w n t h a t t h e q u e s t i o n i s w r o n g!$ $j u s t p u t x = 0 o r x = 1!$ $y o u c a n n o t p r o v e s o m e t h i n g w h i c h$ $i s n o t t r u e!$
Commented by BaliramKumar last updated on 01/May/23

$you mean find the value of x ?$
Commented by mathlove last updated on 02/May/23

$y e s$
	Answered by BaliramKumar last updated on 01/May/23

	$\frac{2 x}{2 \times 3 \times 4} - \frac{4}{2 \times 3 \times 4} + \frac{3 x}{3 \times 4 \times 5} - \frac{5}{3 \times 4 \times 5} + . . . . . . . . . . . + \frac{100 x}{100 \times 101 \times 102} - \frac{102}{100 \times 101 \times 102} = \frac{103}{102}$ $\frac{x}{3 \times 4} - \frac{1}{2 \times 3} + \frac{x}{4 \times 5} - \frac{1}{3 \times 4} + . . . . . . . . . . . + \frac{x}{101 \times 102} - \frac{1}{100 \times 101} = \frac{103}{102}$ $\frac{x}{3 \times 4} - \frac{1}{2 \times 3} + \frac{x}{4 \times 5} - \frac{1}{3 \times 4} + . . . . . . . . . . . + \frac{x}{101 \times 102} - \frac{1}{100 \times 101} = \frac{103}{102}$ $x [\frac{1}{3} - \frac{1}{102}] - [\frac{1}{2} - \frac{1}{101}] = \frac{103}{102}$ $x [\frac{34 - 1}{102}] - [\frac{101 - 2}{2 \times 101}] = \frac{103}{102}$ $x [\frac{33}{102}] - [\frac{99}{202}] = \frac{103}{102}$ $x [\frac{11}{34}] = \frac{103}{102} + \frac{99}{202}$ $\frac{11 x}{34} = \frac{7726}{5151}$ $x = \frac{15452}{3333}$
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