Find the sum of the nth term of the series: (1/(1.2.3)) + (1/(4.5.6)) + (1/(7.8.9)) + ...


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Question Number 44986 by Tawa1 last updated on 07/Oct/18

$Find the sum of the nth term of the series : \frac{1}{1.2.3} + \frac{1}{4.5.6} + \frac{1}{7.8.9} + . . .$
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Oct/18

$p l s w r i t e t h e a n s w e r o f q u e s t i i n s s n d s o u r c e o f$ $q u e s t i o n . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 07/Oct/18
	$T_n =(1/({1+(n−1)×3}(2+(n−1)×3}{3+(n−1)×3})) T_n =(1/((3n−2)(3n−1)(3n))) here i am using formula tl be attached later S_n =C−(1/((3n−1)3n×1×2)) (1/(1×2×3))=C−(1/(2×3×2)) C=(1/6)+(1/(12))=((2+1)/(12))=(1/4) S_n =(1/4)−(1/(2(3n−1)(3n))) Pls check recheck... S_2 =(1/4)−(1/(2×5×6))=((15−1)/(60))=(7/(30)) but (1/(1×2×3))+(1/(4×5×6))=(1/6)+(1/(120))=((20+1)/(120))=(7/(40)) n=3 S_3 =(1/4)−(1/(2(8)(9)))=((36−1)/(144))=((35)/(144)) (1/6)+(1/(120))+(1/(210)) some problem is there i can not detect...$
	$T_{n} = \frac{1}{{1 + (n - 1) \times 3} (2 + (n - 1) \times 3} {3 + (n - 1) \times 3}}$ $T_{n} = \frac{1}{(3 n - 2) (3 n - 1) (3 n)}$ $h e r e i a m u s i n g f o r m u l a t l b e a t t a c h e d l a t e r$ $S_{n} = C - \frac{1}{(3 n - 1) 3 n \times 1 \times 2}$ $\frac{1}{1 \times 2 \times 3} = C - \frac{1}{2 \times 3 \times 2}$ $C = \frac{1}{6} + \frac{1}{12} = \frac{2 + 1}{12} = \frac{1}{4}$ $S_{n} = \frac{1}{4} - \frac{1}{2 (3 n - 1) (3 n)}$ $P l s c h e c k$ $r e c h e c k . . .$ $S_{2} = \frac{1}{4} - \frac{1}{2 \times 5 \times 6} = \frac{15 - 1}{60} = \frac{7}{30}$ $b u t \frac{1}{1 \times 2 \times 3} + \frac{1}{4 \times 5 \times 6} = \frac{1}{6} + \frac{1}{120} = \frac{20 + 1}{120} = \frac{7}{40}$ $n = 3$ $S_{3} = \frac{1}{4} - \frac{1}{2 (8) (9)} = \frac{36 - 1}{144} = \frac{35}{144}$ $\frac{1}{6} + \frac{1}{120} + \frac{1}{210} s o m e p r o b l e m i s t h e r e i c a n n o t d e t e c t . . .$
	Commented by Tawa1 last updated on 07/Oct/18

	$No answer there sir .$ $Thanks for everytime sir . God bless you .$
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