(1/(1.3.5))+(1/(3.5.7))+(1/(5.7.9))+.....nth term


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Question Number 46245 by rpseelam last updated on 23/Oct/18

$\frac{1}{1.3.5} + \frac{1}{3.5.7} + \frac{1}{5.7.9} + . . . . . n t h t e r m$
Commented by maxmathsup by imad last updated on 23/Oct/18
$let S_n =Σ_(k=0) ^n (1/((2k+1)(2k+3)(2k+5))) let find S_n let decompose F(x)=(1/((2x+1)(2x+3)(2x+5))) ⇒F(x)=(a/(2x+1)) +(b/(2x+3)) +(c/(2x+5)) a =lim_(x→−(1/2)) (2x+1)F(x)= (1/(2.4)) =(1/8) b=lim_(x→−(3/2)) (2x+3)F(x)= (1/((−2).2)) =−(1/4) c=lim_(x→−(5/2)) (2x+5)F(x)= (1/((−4).(−2))) =(1/8) ⇒ F(x)= (1/(8(2x+1))) −(1/(4(2x+3))) +(1/(8(2x+5))) ⇒ S_n =(1/8)Σ_(k=0) ^n (1/(2k+1)) −(1/4) Σ_(k=0) ^n (1/(2k+3)) +(1/8) Σ_(k=0) ^n (1/(2k+5)) but Σ_(k=0) ^n (1/(2k+3)) =_(k=j−1) Σ_(j=1) ^(n+1) (1/(2j+1)) =Σ_(j=0) ^n (1/(2j+1)) −1 +(1/(2n+1)) Σ_(k=0) ^n (1/(2k+5)) =_(k=j−2) Σ_(j=2) ^(n+2) (1/(2j +1)) =Σ_(j=0) ^n (1/(2j+1)) −1−(1/3) +(1/(2n+1)) +(1/(2n+5)) S_n =(1/8) Σ_(k=0) ^n (1/(2k+1)) −(1/4) Σ_(j=0) ^n (1/(2j+1)) +(1/4) −(1/(4(2n+1))) +(1/8)Σ_(j=0) ^n (1/(2j+1)) +(1/8)(−(4/3)) +(1/(8(2n+1))) +(1/(8(2n+5))) =(1/4) −(1/6) −(1/(4(2n+1))) +(1/(8(2n+))) +(1/(8(2n+5))) =(1/(12)) −(1/(8(2n+1))) +(1/(8(2n+5))) =(1/(12)) +(1/8){(1/(2n+5)) −(1/(2n+1))} =(1/(12)) +(1/8){((−4)/((2n+1)(2n+5)))} =(1/(12)) −(1/(2(2n+1)(2n+5))) .$
$l e t S_{n} = \sum_{k = 0}^{n} \frac{1}{(2 k + 1) (2 k + 3) (2 k + 5)} l e t f i n d S_{n} l e t d e c o m p o s e$ $F (x) = \frac{1}{(2 x + 1) (2 x + 3) (2 x + 5)} \Rightarrow F (x) = \frac{a}{2 x + 1} + \frac{b}{2 x + 3} + \frac{c}{2 x + 5}$ $a = {l i m}_{x \to - \frac{1}{2}} (2 x + 1) F (x) = \frac{1}{2.4} = \frac{1}{8}$ $b = {l i m}_{x \to - \frac{3}{2}} (2 x + 3) F (x) = \frac{1}{(- 2) . 2} = - \frac{1}{4}$ $c = {l i m}_{x \to - \frac{5}{2}} (2 x + 5) F (x) = \frac{1}{(- 4) . (- 2)} = \frac{1}{8} \Rightarrow$ $F (x) = \frac{1}{8 (2 x + 1)} - \frac{1}{4 (2 x + 3)} + \frac{1}{8 (2 x + 5)} \Rightarrow$ $S_{n} = \frac{1}{8} \sum_{k = 0}^{n} \frac{1}{2 k + 1} - \frac{1}{4} \sum_{k = 0}^{n} \frac{1}{2 k + 3} + \frac{1}{8} \sum_{k = 0}^{n} \frac{1}{2 k + 5} b u t$ $\sum_{k = 0}^{n} \frac{1}{2 k + 3} =_{k = j - 1} \sum_{j = 1}^{n + 1} \frac{1}{2 j + 1} = \sum_{j = 0}^{n} \frac{1}{2 j + 1} - 1 + \frac{1}{2 n + 1}$ $\sum_{k = 0}^{n} \frac{1}{2 k + 5} =_{k = j - 2} \sum_{j = 2}^{n + 2} \frac{1}{2 j + 1} = \sum_{j = 0}^{n} \frac{1}{2 j + 1} - 1 - \frac{1}{3} + \frac{1}{2 n + 1} + \frac{1}{2 n + 5}$ $S_{n} = \frac{1}{8} \sum_{k = 0}^{n} \frac{1}{2 k + 1} - \frac{1}{4} \sum_{j = 0}^{n} \frac{1}{2 j + 1} + \frac{1}{4} - \frac{1}{4 (2 n + 1)}$ $+ \frac{1}{8} \sum_{j = 0}^{n} \frac{1}{2 j + 1} + \frac{1}{8} (- \frac{4}{3}) + \frac{1}{8 (2 n + 1)} + \frac{1}{8 (2 n + 5)}$ $= \frac{1}{4} - \frac{1}{6} - \frac{1}{4 (2 n + 1)} + \frac{1}{8 (2 n +)} + \frac{1}{8 (2 n + 5)}$ $= \frac{1}{12} - \frac{1}{8 (2 n + 1)} + \frac{1}{8 (2 n + 5)} = \frac{1}{12} + \frac{1}{8} {\frac{1}{2 n + 5} - \frac{1}{2 n + 1}}$ $= \frac{1}{12} + \frac{1}{8} {\frac{- 4}{(2 n + 1) (2 n + 5)}} = \frac{1}{12} - \frac{1}{2 (2 n + 1) (2 n + 5)} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18
	$T_n =(1/({1+(n−1)2}{3+(n−1)2}{5+(n−1)2})) T_n =(1/((2n−1)(2n+1)(2n+3))) S_n =C−(1/((2n+1)(2n+3)×2×2)) put n=1 S_1 =T_1 =(1/(15)) (1/(15))=C−(1/(4(3)(5))) C=(1/(15))+(1/(60))=((4+1)/(60))=(1/(12)) S_n =(1/(12))−(1/(4(2n+1)(2n+3))) checking the answer by follosing proof 1)S_1 =T_1 2)S_2 =T_1 +T_2 3)S_3 =T_1 +T_2 +T_3 T_n =(1/((2n−1)(2n+1)((2n+3))) T_1 =(1/(1×3×5))=(1/(15)) S_1 =(1/(12))−(1/(4×3×5))=(1/(15)) T_1 =S_1 proved T_2 =(1/(3×5×7))=(1/(105)) T_3 =(1/(5×7×9))=(1/(315)) T_1 +T_2 =(1/(15))+(1/(105))=(8/(105)) S_n =(1/(12))−(1/(4(2n+1)(2n+3))) S_2 =(1/(12))−(1/(4×5×7))=(8/(105)) so S_2 =T_1 +T_2 proved T_1 +T_2 +T_3 =(8/(105))+(1/(315))=(5/(63)) S_3 =(1/(12))−(1/(4×7×9))=(5/(63)) so my answer iscorrect...$
	$T_{n} = \frac{1}{{1 + (n - 1) 2} {3 + (n - 1) 2} {5 + (n - 1) 2}}$ $T_{n} = \frac{1}{(2 n - 1) (2 n + 1) (2 n + 3)}$ $S_{n} = C - \frac{1}{(2 n + 1) (2 n + 3) \times 2 \times 2}$ $p u t n = 1 S_{1} = T_{1} = \frac{1}{15}$ $\frac{1}{15} = C - \frac{1}{4 (3) (5)}$ $C = \frac{1}{15} + \frac{1}{60} = \frac{4 + 1}{60} = \frac{1}{12}$ $S_{n} = \frac{1}{12} - \frac{1}{4 (2 n + 1) (2 n + 3)}$ $c h e c k i n g t h e a n s w e r b y f o l l o s i n g p r o o f$ $1) S_{1} = T_{1}$ $2) S_{2} = T_{1} + T_{2}$ $3) S_{3} = T_{1} + T_{2} + T_{3}$ $T_{n} = \frac{1}{(2 n - 1) (2 n + 1) ((2 n + 3)}$ $T_{1} = \frac{1}{1 \times 3 \times 5} = \frac{1}{15} S_{1} = \frac{1}{12} - \frac{1}{4 \times 3 \times 5} = \frac{1}{15} T_{1} = S_{1} p r o v e d$ $T_{2} = \frac{1}{3 \times 5 \times 7} = \frac{1}{105}$ $T_{3} = \frac{1}{5 \times 7 \times 9} = \frac{1}{315}$ $T_{1} + T_{2} = \frac{1}{15} + \frac{1}{105} = \frac{8}{105}$ $S_{n} = \frac{1}{12} - \frac{1}{4 (2 n + 1) (2 n + 3)}$ $S_{2} = \frac{1}{12} - \frac{1}{4 \times 5 \times 7} = \frac{8}{105} s o S_{2} = T_{1} + T_{2} p r o v e d$ $T_{1} + T_{2} + T_{3} = \frac{8}{105} + \frac{1}{315} = \frac{5}{63}$ $S_{3} = \frac{1}{12} - \frac{1}{4 \times 7 \times 9} = \frac{5}{63}$ $s o m y a n s w e r i s c o r r e c t . . .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 23/Oct/18

	$m y a n s i s c o r r e c t . . .$
	Commented by rpseelam last updated on 24/Oct/18

	$t h a t i s e x c e l l e n t s i r . t h a n k s$
	Commented by tanmay.chaudhury50@gmail.com last updated on 24/Oct/18

	$m o s t w e l c o m e . . .$
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