I heard this sum can be close using the Digamma function. Please help me use it. i don′t know it. sum of nth term: (1/(1.2.3)) + (1/(4.5.6)) + (1/(7.8.9)) + ...


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Question Number 45316 by Tawa1 last updated on 11/Oct/18

$I heard this sum can be close using the Digamma function .$ $Please help me use it . i {don}^{'} t know it .$ $sum of nth term : \frac{1}{1.2.3} + \frac{1}{4.5.6} + \frac{1}{7.8.9} + . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18
	$T_n =(1/({1+(n−1)3}{2+(n−1)3}{3+(n−1)3})) T_n =(1/((3n−2)((3n−1)(3n))) here i am using some formula mentioned Hall and knight Algebra... S_n =C−(1/((3n−1)(3n)×2×1)) n=1 (1/(1×2×3))=C−(1/(2×3×2)) C=(1/6)+(1/(12))=((2+1)/(12))=(1/4) S_n =(1/4)−(1/(2(3n−1)(3n))) recheck... S_1 =(1/4)−(1/(2×2×3))=((3−1)/(12))=(1/6)=T_1 =(1/(1×2×3))=(1/6) conform S_2 =(1/4)−(1/(2×5×6))=((15−1)/(60))=((14)/(60))=(7/(30)) T_2 =S_2 −S_1 =(7/(30))−(1/6)=((7−5)/(30))=(1/(15)) but T_2 =(1/(120)) So my answer is not correct... pls menhion the source of this problem S_n =(1/(1×2×3))+(1/(4×5×6))+(1/(7×8×9))+... =$
	$T_{n} = \frac{1}{{1 + (n - 1) 3} {2 + (n - 1) 3} {3 + (n - 1) 3}}$ $T_{n} = \frac{1}{(3 n - 2) ((3 n - 1) (3 n)}$ $h e r e i a m u s i n g s o m e f o r m u l a m e n t i o n e d$ $H a l l a n d k n i g h t A l g e b r a . . .$ $S_{n} = C - \frac{1}{(3 n - 1) (3 n) \times 2 \times 1}$ $n = 1$ $\frac{1}{1 \times 2 \times 3} = C - \frac{1}{2 \times 3 \times 2}$ $C = \frac{1}{6} + \frac{1}{12} = \frac{2 + 1}{12} = \frac{1}{4}$ $S_{n} = \frac{1}{4} - \frac{1}{2 (3 n - 1) (3 n)}$ $r e c h e c k . . .$ $S_{1} = \frac{1}{4} - \frac{1}{2 \times 2 \times 3} = \frac{3 - 1}{12} = \frac{1}{6} = T_{1} = \frac{1}{1 \times 2 \times 3} = \frac{1}{6}$ $c o n f o r m$ $S_{2} = \frac{1}{4} - \frac{1}{2 \times 5 \times 6} = \frac{15 - 1}{60} = \frac{14}{60} = \frac{7}{30}$ $T_{2} = S_{2} - S_{1} = \frac{7}{30} - \frac{1}{6} = \frac{7 - 5}{30} = \frac{1}{15}$ $b u t T_{2} = \frac{1}{120}$ $S o m y a n s w e r i s n o t c o r r e c t . . .$ $p l s m e n h i o n t h e s o u r c e o f t h i s p r o b l e m$ $S_{n} = \frac{1}{1 \times 2 \times 3} + \frac{1}{4 \times 5 \times 6} + \frac{1}{7 \times 8 \times 9} + . . .$ $=$
	Commented by Tawa1 last updated on 12/Oct/18

	$The question was asked online .$ $But sir, when can i apply the formular you used to find the sum .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

	$o k . . i a m p o s t i n g . . .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 12/Oct/18

	Commented by Tawa1 last updated on 12/Oct/18

	$Wow, God bless you sir . I really appreciate your effort .$
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