Question Number 202251 by BaliramKumar last updated on 23/Dec/23
$$\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}×\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}×\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{3}×\mathrm{4}×\mathrm{5}}\:+\:…………..\:+\:\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)}\:=\:?\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Commented by BaliramKumar last updated on 23/Dec/23
$$\frac{\mathrm{1}}{\mathrm{a}\left(\mathrm{a}+\mathrm{d}\right)\left(\mathrm{a}+\mathrm{2d}\right)}\:+\:\frac{\mathrm{1}}{\left(\mathrm{a}+\mathrm{d}\right)\left(\mathrm{a}+\mathrm{2d}\right)\left(\mathrm{a}+\mathrm{3d}\right)}\:+\:\:…….+\:\frac{\mathrm{1}}{\left\{\mathrm{a}+\left(\mathrm{n}−\mathrm{1}\right)\mathrm{d}\right\}\left\{\mathrm{a}+\mathrm{nd}\right\}\left\{\mathrm{a}+\left(\mathrm{n}+\mathrm{1}\right)\mathrm{d}\right\}}\:=\:? \\ $$
Commented by MATHEMATICSAM last updated on 24/Dec/23
![T_1 = (1/(a(a + d)(a + 2d))) , T_2 = (1/((a + d)(a + 2d)(a + 3d))) , ........ T_(n ) = (1/({a + (n − 1)d}{a + nd}{a + (n + 1)d})) = (1/(2d))[(({a + (n + 1)d }− {a + (n − 1)d})/({a + (n − 1)d}{a + nd}{a + (n + 1)d}))] = (1/(2d))[(1/({a + (n − 1)d}{a + nd})) − (1/({a + nd}{a + (n + 1)d}))] T_1 = (1/(2d))[(1/(a(a + d))) − (1/((a + d)(a + 2d)))] T_2 = (1/(2d))[(1/((a + d)(a + 2d))) − (1/((a + 2d)(a + 3d)))] T_3 = (1/(2d))[(1/((a + 2d)(a + 3d))) − (1/((a + 3d)(a + 4d)))] . . . . T_(n − 1) = (1/(2d))[(1/({a + (n − 2)d}{a + (n − 1)d})) − (1/({a + (n − 1)d}{a + nd}))] T_n = (1/(2d))[(1/({a + (n − 1)d}{a + nd})) − (1/({a + nd}{a + (n + 1)d}))] Now T_1 + T_2 + T_3 + .... + T_n = (1/(2d))[(1/(a(a + d))) − (1/({a + nd}{a + (n + 1)d}))]](https://www.tinkutara.com/question/Q202283.png)
$$\mathrm{T}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{{a}\left({a}\:+\:{d}\right)\left({a}\:+\:\mathrm{2}{d}\right)}\:, \\ $$$$\mathrm{T}_{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\left({a}\:+\:{d}\right)\left({a}\:+\:\mathrm{2}{d}\right)\left({a}\:+\:\mathrm{3}{d}\right)}\:,\:……..\:\: \\ $$$$\mathrm{T}_{{n}\:} =\:\frac{\mathrm{1}}{\left\{{a}\:+\:\left({n}\:−\:\mathrm{1}\right){d}\right\}\left\{{a}\:+\:{nd}\right\}\left\{{a}\:+\:\left({n}\:+\:\mathrm{1}\right){d}\right\}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}{d}}\left[\frac{\left\{{a}\:+\:\left({n}\:+\:\mathrm{1}\right){d}\:\right\}−\:\left\{{a}\:+\:\left({n}\:−\:\mathrm{1}\right){d}\right\}}{\left\{{a}\:+\:\left({n}\:−\:\mathrm{1}\right){d}\right\}\left\{{a}\:+\:{nd}\right\}\left\{{a}\:+\:\left({n}\:+\:\mathrm{1}\right){d}\right\}}\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}{d}}\left[\frac{\mathrm{1}}{\left\{{a}\:+\:\left({n}\:−\:\mathrm{1}\right){d}\right\}\left\{{a}\:+\:{nd}\right\}}\:−\:\frac{\mathrm{1}}{\left\{{a}\:+\:{nd}\right\}\left\{{a}\:+\:\left({n}\:+\:\mathrm{1}\right){d}\right\}}\right] \\ $$$$ \\ $$$$ \\ $$$$\mathrm{T}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}{d}}\left[\frac{\mathrm{1}}{{a}\left({a}\:+\:{d}\right)}\:−\:\cancel{\frac{\mathrm{1}}{\left({a}\:+\:{d}\right)\left({a}\:+\:\mathrm{2}{d}\right)}}\right] \\ $$$$\mathrm{T}_{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}{d}}\left[\cancel{\frac{\mathrm{1}}{\left({a}\:+\:{d}\right)\left({a}\:+\:\mathrm{2}{d}\right)}\:}−\:\cancel{\frac{\mathrm{1}}{\left({a}\:+\:\mathrm{2}{d}\right)\left({a}\:+\:\mathrm{3}{d}\right)}}\right] \\ $$$$\mathrm{T}_{\mathrm{3}} \:=\:\frac{\mathrm{1}}{\mathrm{2}{d}}\left[\cancel{\frac{\mathrm{1}}{\left({a}\:+\:\mathrm{2}{d}\right)\left({a}\:+\:\mathrm{3}{d}\right)}}\:−\:\cancel{\frac{\mathrm{1}}{\left({a}\:+\:\mathrm{3}{d}\right)\left({a}\:+\:\mathrm{4}{d}\right)}}\right] \\ $$$$. \\ $$$$. \\ $$$$. \\ $$$$. \\ $$$$\mathrm{T}_{{n}\:−\:\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}{d}}\left[\cancel{\frac{\mathrm{1}}{\left\{{a}\:+\:\left({n}\:−\:\mathrm{2}\right){d}\right\}\left\{{a}\:+\:\left({n}\:−\:\mathrm{1}\right){d}\right\}}}\:−\:\cancel{\frac{\mathrm{1}}{\left\{{a}\:+\:\left({n}\:−\:\mathrm{1}\right){d}\right\}\left\{{a}\:+\:{nd}\right\}}}\right] \\ $$$$\mathrm{T}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{2}{d}}\left[\cancel{\frac{\mathrm{1}}{\left\{{a}\:+\:\left({n}\:−\:\mathrm{1}\right){d}\right\}\left\{{a}\:+\:{nd}\right\}}}\:−\:\frac{\mathrm{1}}{\left\{{a}\:+\:{nd}\right\}\left\{{a}\:+\:\left({n}\:+\:\mathrm{1}\right){d}\right\}}\right] \\ $$$$ \\ $$$$\mathrm{Now}\:\mathrm{T}_{\mathrm{1}} \:+\:\mathrm{T}_{\mathrm{2}} \:+\:\mathrm{T}_{\mathrm{3}} \:+\:….\:+\:\mathrm{T}_{{n}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}{d}}\left[\frac{\mathrm{1}}{{a}\left({a}\:+\:{d}\right)}\:−\:\frac{\mathrm{1}}{\left\{{a}\:+\:{nd}\right\}\left\{{a}\:+\:\left({n}\:+\:\mathrm{1}\right){d}\right\}}\right] \\ $$
Commented by BaliramKumar last updated on 24/Dec/23
![Thanks Sir we can also write in this form = (n/2)[((a+{a+(n+1)d})/(a(a + d){a+nd}{a+(n+1)d}))]](https://www.tinkutara.com/question/Q202284.png)
$$\mathrm{Thanks}\:\mathrm{Sir} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{also}\:\mathrm{write}\:\mathrm{in}\:\mathrm{this}\:\mathrm{form} \\ $$$$=\:\frac{\mathrm{n}}{\mathrm{2}}\left[\frac{\mathrm{a}+\left\{\mathrm{a}+\left(\mathrm{n}+\mathrm{1}\right)\mathrm{d}\right\}}{{a}\left({a}\:+\:{d}\right)\left\{\mathrm{a}+\mathrm{nd}\right\}\left\{\mathrm{a}+\left(\mathrm{n}+\mathrm{1}\right)\mathrm{d}\right\}}\right] \\ $$$$ \\ $$
Commented by MATHEMATICSAM last updated on 24/Dec/23
$$\mathrm{yes} \\ $$
Answered by MATHEMATICSAM last updated on 23/Dec/23
![Let s = (1/(1×2×3)) + (1/(2×3×4)) + (1/(3×4×5)) + .... + (1/(n(n+1)(n+2))) T_1 = (1/(1× 2 × 3)) , T_2 = (1/(2 × 3 × 4)) , T_3 = (1/(3 × 4 × 5)) , ....... T_n = (1/(n(n + 1)(n + 2))) = (2/(n(n + 1)(n + 2))) × (1/2) = (((n + 2) − n)/(n(n + 1)(n + 2))) × (1/2) = (1/2)[(1/(n(n + 1))) − (1/((n + 1)(n + 2)))] T_1 = (1/2)[(1/(1 × 2)) − (1/(2 × 3))] T_2 = (1/2)[(1/(2 × 3)) − (1/(3 × 4))] T_3 = (1/2)[(1/(3 × 4)) − (1/(4 × 5))] . . . . T_(n − 1) = (1/2)[(1/((n − 1)n)) − (1/(n(n + 1)))] T_n = (1/2)[(1/(n(n + 1))) − (1/((n + 1)(n + 2)))] Now s = T_1 + T_2 + T_3 + .... + T_(n ) = (1/2)[(1/2) − (1/((n + 1)(n + 2)))] = (1/2)[(((n + 1)(n + 2) − 2)/(2(n + 1)(n + 2)))] = ((n^2 + 3n + 2 − 2 )/(4(n + 1)(n + 2))) = ((n(n + 3))/(4(n + 1)(n + 2))) (Ans)](https://www.tinkutara.com/question/Q202253.png)
$$\mathrm{Let}\:{s}\:=\:\frac{\mathrm{1}}{\mathrm{1}×\mathrm{2}×\mathrm{3}}\:+\:\frac{\mathrm{1}}{\mathrm{2}×\mathrm{3}×\mathrm{4}}\:+\:\frac{\mathrm{1}}{\mathrm{3}×\mathrm{4}×\mathrm{5}}\:+\:….\:+\:\frac{\mathrm{1}}{\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\left(\mathrm{n}+\mathrm{2}\right)} \\ $$$$\mathrm{T}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{1}×\:\mathrm{2}\:×\:\mathrm{3}}\:,\:\mathrm{T}_{\mathrm{2}} \:=\:\:\frac{\mathrm{1}}{\mathrm{2}\:×\:\mathrm{3}\:×\:\mathrm{4}}\:, \\ $$$$\mathrm{T}_{\mathrm{3}} \:=\:\frac{\mathrm{1}}{\mathrm{3}\:×\:\mathrm{4}\:×\:\mathrm{5}}\:,\:……. \\ $$$$\mathrm{T}_{{n}} \:=\:\frac{\mathrm{1}}{{n}\left({n}\:+\:\mathrm{1}\right)\left({n}\:+\:\mathrm{2}\right)}\:=\:\frac{\mathrm{2}}{{n}\left({n}\:+\:\mathrm{1}\right)\left({n}\:+\:\mathrm{2}\right)}\:×\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:=\:\frac{\left({n}\:+\:\mathrm{2}\right)\:−\:{n}}{{n}\left({n}\:+\:\mathrm{1}\right)\left({n}\:+\:\mathrm{2}\right)}\:×\:\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{{n}\left({n}\:+\:\mathrm{1}\right)}\:−\:\frac{\mathrm{1}}{\left({n}\:+\:\mathrm{1}\right)\left({n}\:+\:\mathrm{2}\right)}\right] \\ $$$$\mathrm{T}_{\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{1}\:×\:\mathrm{2}}\:−\:\cancel{\frac{\mathrm{1}}{\mathrm{2}\:×\:\mathrm{3}}}\right] \\ $$$$\mathrm{T}_{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\cancel{\frac{\mathrm{1}}{\mathrm{2}\:×\:\mathrm{3}}}\:−\:\cancel{\frac{\mathrm{1}}{\mathrm{3}\:×\:\mathrm{4}}}\right] \\ $$$$\mathrm{T}_{\mathrm{3}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\cancel{\frac{\mathrm{1}}{\mathrm{3}\:×\:\mathrm{4}}}\:−\:\cancel{\frac{\mathrm{1}}{\mathrm{4}\:×\:\mathrm{5}}}\right] \\ $$$$. \\ $$$$. \\ $$$$. \\ $$$$. \\ $$$$\mathrm{T}_{{n}\:−\:\mathrm{1}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\cancel{\frac{\mathrm{1}}{\left({n}\:−\:\mathrm{1}\right){n}}}\:−\cancel{\:\frac{\mathrm{1}}{{n}\left({n}\:+\:\mathrm{1}\right)}}\right] \\ $$$$\mathrm{T}_{{n}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\cancel{\frac{\mathrm{1}}{{n}\left({n}\:+\:\mathrm{1}\right)}}\:−\:\frac{\mathrm{1}}{\left({n}\:+\:\mathrm{1}\right)\left({n}\:+\:\mathrm{2}\right)}\right] \\ $$$$ \\ $$$$\mathrm{Now}\: \\ $$$${s}\:=\:\mathrm{T}_{\mathrm{1}} \:+\:\mathrm{T}_{\mathrm{2}} \:+\:\mathrm{T}_{\mathrm{3}} \:+\:….\:+\:\mathrm{T}_{{n}\:} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{2}}\:−\:\frac{\mathrm{1}}{\left({n}\:+\:\mathrm{1}\right)\left({n}\:+\:\mathrm{2}\right)}\right] \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\left({n}\:+\:\mathrm{1}\right)\left({n}\:+\:\mathrm{2}\right)\:−\:\mathrm{2}}{\mathrm{2}\left({n}\:+\:\mathrm{1}\right)\left({n}\:+\:\mathrm{2}\right)}\right] \\ $$$$=\:\frac{{n}^{\mathrm{2}} \:+\:\mathrm{3}{n}\:+\:\mathrm{2}\:−\:\mathrm{2}\:}{\mathrm{4}\left({n}\:+\:\mathrm{1}\right)\left({n}\:+\:\mathrm{2}\right)} \\ $$$$=\:\frac{{n}\left({n}\:+\:\mathrm{3}\right)}{\mathrm{4}\left({n}\:+\:\mathrm{1}\right)\left({n}\:+\:\mathrm{2}\right)}\:\left(\boldsymbol{\mathrm{Ans}}\right) \\ $$
Commented by BaliramKumar last updated on 23/Dec/23
$$\mathrm{Thanks}\:\mathrm{Sir} \\ $$