Find-the-sum-of-the-nth-term-of-the-series-1-1-2-3-1-4-5-6-1-7-8-9-

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} + \dots

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 \dots

Answered by tanmay.chaudhury50@gmail.com last updated on 07/Oct/18

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 \dots

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 \dots

Commented by Tawa1 last updated on 07/Oct/18

No answer there sir .

Thanks for everytime sir . God bless you .

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