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Question Number 202251 by BaliramKumar last updated on 23/Dec/23

(1/(1×2×3)) + (1/(2×3×4)) + (1/(3×4×5)) + .............. + (1/(n(n+1)(n+2))) = ?

11×2×3+12×3×4+13×4×5+..............+1n(n+1)(n+2)=?

Commented by BaliramKumar last updated on 23/Dec/23

(1/(a(a+d)(a+2d))) + (1/((a+d)(a+2d)(a+3d))) +  .......+ (1/({a+(n−1)d}{a+nd}{a+(n+1)d})) = ?

1a(a+d)(a+2d)+1(a+d)(a+2d)(a+3d)+.......+1{a+(n1)d}{a+nd}{a+(n+1)d}=?

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}))]

T1=1a(a+d)(a+2d),T2=1(a+d)(a+2d)(a+3d),........Tn=1{a+(n1)d}{a+nd}{a+(n+1)d}=12d[{a+(n+1)d}{a+(n1)d}{a+(n1)d}{a+nd}{a+(n+1)d}]=12d[1{a+(n1)d}{a+nd}1{a+nd}{a+(n+1)d}]T1=12d[1a(a+d)1(a+d)(a+2d)]T2=12d[1(a+d)(a+2d)1(a+2d)(a+3d)]T3=12d[1(a+2d)(a+3d)1(a+3d)(a+4d)]....Tn1=12d[1{a+(n2)d}{a+(n1)d}1{a+(n1)d}{a+nd}]Tn=12d[1{a+(n1)d}{a+nd}1{a+nd}{a+(n+1)d}]NowT1+T2+T3+....+Tn=12d[1a(a+d)1{a+nd}{a+(n+1)d}]

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}))]

ThanksSirwecanalsowriteinthisform=n2[a+{a+(n+1)d}a(a+d){a+nd}{a+(n+1)d}]

Commented by MATHEMATICSAM last updated on 24/Dec/23

yes

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)

Lets=11×2×3+12×3×4+13×4×5+....+1n(n+1)(n+2)T1=11×2×3,T2=12×3×4,T3=13×4×5,.......Tn=1n(n+1)(n+2)=2n(n+1)(n+2)×12=(n+2)nn(n+1)(n+2)×12=12[1n(n+1)1(n+1)(n+2)]T1=12[11×212×3]T2=12[12×313×4]T3=12[13×414×5]....Tn1=12[1(n1)n1n(n+1)]Tn=12[1n(n+1)1(n+1)(n+2)]Nows=T1+T2+T3+....+Tn=12[121(n+1)(n+2)]=12[(n+1)(n+2)22(n+1)(n+2)]=n2+3n+224(n+1)(n+2)=n(n+3)4(n+1)(n+2)(Ans)

Commented by BaliramKumar last updated on 23/Dec/23

Thanks Sir

ThanksSir

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