Question Number 100223 by DGmichael last updated on 25/Jun/20

Answered by maths mind last updated on 25/Jun/20

A=Σ_(k=0) ^n (((−1)^k C_n ^k )/(k+1))  let f(x)=(1−x)^n ⇒f(x)=Σ_(k=0) ^n C_n ^k (−1)^k x^k   ∫_0 ^1 (1−x)^n dx=∫_0 ^1 Σ_(k=0) ^n C_n ^k (−1)^k x^k dx  =Σ_(k=0) ^n C_n ^k (−1)^k ∫_0 ^1 x^k dx=ΣC_n ^k (−1)^k .(1/(k+1))  =Σ_(k=0) ^n (((−1)^k C_n ^k )/(k+1))=∫_0 ^1 (1−x)^n dx=(1/(1+n))

Commented byAr Brandon last updated on 25/Jun/20

wow amazing ! you quickly recognized the integral.😃

Commented byAr Brandon last updated on 25/Jun/20

Excuse me Sir, Is there any theory for this, or any particular topic that deals with these kind of sums ?

Commented bymaths mind last updated on 25/Jun/20

the main idea  is newtoon identitie  (x+y)^n =Σ_(k=0) ^n C_n ^k x^k y^(n−k)   then use it integral or differebtial  ∂_x (x+y)^n =∂_x Σ_(k=0) ^n C_n ^k x^k y^(n−k) ⇔n(x+y)^(n−1) =Σ_(k≥1) C_n ^k kx^(k−1) y^(n−k)   or (((x+y)^(n+1) )/(n+1))=ΣC_n ^k ((x^(k+1) y^(n−k) )/(k+1))  exempl   we want  Σ_(k=0) ^n (C_n ^k /((k+1)(k+2)(k+3)))...  (1+x)^n =Σ_(k=0) ^n C_n ^k x^k ⇒∫_0 ^t (1+x)^n dx=Σ∫_0 ^t C_n ^k x^k dx  ⇒(((1+t)^n −1)/(n+1))=Σ_(k=0) ^n ((C_n ^k t^(k+1) )/(k+1))  ∫_0 ^x (((1+t)^n −1)/((n+1)))dt=Σ_(k=0) ^n C_n ^k ∫_0 ^x (t^(k+1) /((k+1)))dt=ΣC_n ^k (x^(k+2) /((k+1)(k+2)))  (((1+x)^(n+2) )/((n+1)(n+2)))−(x/(n+1))=ΣC_n ^k (x^(k+2) /((k+1)(k+2)))  ⇒∫_0 ^y (((1+x)^(n+2) )/((n+1)(n+2)))−(x/(n+1))=ΣC_n ^k ∫_0 ^y (x^(k+2) /((k+1)(k+2)))  ⇔(((1+y)^(n+3) )/((n+1)(n+2)(n+3)))−(1/((n+1)(n+2)))−(y^2 /(2(n+1)))=Σ_(k=0) ^n C_n ^k (y^(k+3) /((k+2)(k+1)(k+3)))  y=1⇔(2^(n+3) /((n+1)(n+2)(n+3)))−(1/((n+1)(n+2)))−(1/(2(n+1)))=Σ_(k=0) ^n C_n ^k .(1/((k+1)(k+2)(k+3)))  we can Show  Σ_(k=0) ^n (C_n ^k /(Π_(b=1) ^r (k+b)))=(2^(r+n) /((n+1).....(n+r)))−Σ_(j=1) ^(r−1) (1/((r−j)!Π_(k=1) ^j (n+j)))

Commented byAr Brandon last updated on 26/Jun/20

Thanks a lot for your time Sir💞

Commented byDGmichael last updated on 26/Jun/20

😁 thanks dear sir!