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1-1-3-2-1-3-5-3-1-3-5-7-n-terms-Find-sum-




Question Number 105250 by bshahid010@gmail.com last updated on 27/Jul/20
(1/(1×3))+(2/(1×3×5))+(3/(1×3×5×7))+.........n−terms  Find sum
$$\frac{\mathrm{1}}{\mathrm{1}×\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{1}×\mathrm{3}×\mathrm{5}}+\frac{\mathrm{3}}{\mathrm{1}×\mathrm{3}×\mathrm{5}×\mathrm{7}}+………\mathrm{n}−\mathrm{terms} \\ $$$$\mathrm{Find}\:\mathrm{sum} \\ $$
Answered by nimnim last updated on 27/Jul/20
Let S_n =(1/(1×3))+(2/(1×3×5))+(3/(1×3×5×7))+....to n terms  Nr=n^(th)  term of 1,2,3,.....=n  Dr=product of the first (n+1) terms of            AP  1,3,5....(2n+1),  n∈N  ∴ t_n =(n/(1.3.5.......(2n+1)))=(1/2)[(((2n+1)−1)/(1.3.5....(2n−1)(2n+1)))]            =(1/2)[(1/(1.3.5.....(2n−1)))−(1/(1.3.5.....(2n+1)))]  then  t_1 =(1/2)[(1/1)−(1/(1.3))]  t_2 =(1/2)[(1/(1.3))−(1/(1.3.5))]  t_3 =(1/2)[(1/(1.3.5))−(1/(1.3.5.7))]          ......    .......    .......          ......    .......    .......  t_n =(1/2)[(1/(1.3.5....(2n−1)))−(1/(1.3.5....(2n+1)))]  ∴S_n =(1/2)[1−(1/(1.3.5....(2n+1)))]
$${Let}\:{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{1}×\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{1}×\mathrm{3}×\mathrm{5}}+\frac{\mathrm{3}}{\mathrm{1}×\mathrm{3}×\mathrm{5}×\mathrm{7}}+….{to}\:{n}\:{terms} \\ $$$${Nr}={n}^{{th}} \:{term}\:{of}\:\mathrm{1},\mathrm{2},\mathrm{3},…..={n} \\ $$$${Dr}={product}\:{of}\:{the}\:{first}\:\left({n}+\mathrm{1}\right)\:{terms}\:{of}\: \\ $$$$\:\:\:\:\:\:\:\:\:{AP}\:\:\mathrm{1},\mathrm{3},\mathrm{5}….\left(\mathrm{2}{n}+\mathrm{1}\right),\:\:{n}\in{N} \\ $$$$\therefore\:{t}_{{n}} =\frac{{n}}{\mathrm{1}.\mathrm{3}.\mathrm{5}…….\left(\mathrm{2}{n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)−\mathrm{1}}{\mathrm{1}.\mathrm{3}.\mathrm{5}….\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}.\mathrm{5}…..\left(\mathrm{2}{n}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}.\mathrm{5}…..\left(\mathrm{2}{n}+\mathrm{1}\right)}\right] \\ $$$${then} \\ $$$${t}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}}\right] \\ $$$${t}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}.\mathrm{5}}\right] \\ $$$${t}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}.\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}}\right] \\ $$$$\:\:\:\:\:\:\:\:……\:\:\:\:…….\:\:\:\:……. \\ $$$$\:\:\:\:\:\:\:\:……\:\:\:\:…….\:\:\:\:……. \\ $$$${t}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}.\mathrm{5}….\left(\mathrm{2}{n}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}.\mathrm{5}….\left(\mathrm{2}{n}+\mathrm{1}\right)}\right] \\ $$$$\therefore{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}.\mathrm{5}….\left(\mathrm{2}{n}+\mathrm{1}\right)}\right] \\ $$

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