Question Number 105250 by bshahid010@gmail.com last updated on 27/Jul/20
$$\frac{\mathrm{1}}{\mathrm{1}×\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{1}×\mathrm{3}×\mathrm{5}}+\frac{\mathrm{3}}{\mathrm{1}×\mathrm{3}×\mathrm{5}×\mathrm{7}}+.........\mathrm{n}−\mathrm{terms} \\ $$$$\mathrm{Find}\:\mathrm{sum} \\ $$
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)))]](Q105279.png)
$${Let}\:{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{1}×\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{1}×\mathrm{3}×\mathrm{5}}+\frac{\mathrm{3}}{\mathrm{1}×\mathrm{3}×\mathrm{5}×\mathrm{7}}+....{to}\:{n}\:{terms} \\ $$$${Nr}={n}^{{th}} \:{term}\:{of}\:\mathrm{1},\mathrm{2},\mathrm{3},.....={n} \\ $$$${Dr}={product}\:{of}\:{the}\:{first}\:\left({n}+\mathrm{1}\right)\:{terms}\:{of}\: \\ $$$$\:\:\:\:\:\:\:\:\:{AP}\:\:\mathrm{1},\mathrm{3},\mathrm{5}....\left(\mathrm{2}{n}+\mathrm{1}\right),\:\:{n}\in{N} \\ $$$$\therefore\:{t}_{{n}} =\frac{{n}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.......\left(\mathrm{2}{n}+\mathrm{1}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\left(\mathrm{2}{n}+\mathrm{1}\right)−\mathrm{1}}{\mathrm{1}.\mathrm{3}.\mathrm{5}....\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.....\left(\mathrm{2}{n}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.....\left(\mathrm{2}{n}+\mathrm{1}\right)}\right] \\ $$$${then} \\ $$$${t}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}}\right] \\ $$$${t}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}.\mathrm{5}}\right] \\ $$$${t}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}.\mathrm{5}}−\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}}\right] \\ $$$$\:\:\:\:\:\:\:\:......\:\:\:\:.......\:\:\:\:....... \\ $$$$\:\:\:\:\:\:\:\:......\:\:\:\:.......\:\:\:\:....... \\ $$$${t}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}.\mathrm{5}....\left(\mathrm{2}{n}−\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}.\mathrm{5}....\left(\mathrm{2}{n}+\mathrm{1}\right)}\right] \\ $$$$\therefore{S}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}.\mathrm{5}....\left(\mathrm{2}{n}+\mathrm{1}\right)}\right] \\ $$