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Question Number 52867 by hassentimol last updated on 14/Jan/19

Answered by iv@0uja last updated on 19/Jan/19

     Π_(l=0) ^(m+1) (n+l)−Π_(l=0) ^(m+1) (n+l−1)  =(n+m+1)Π_(l=0) ^m (n+l)−(n−1)Π_(l=0) ^m (n+l)  =((n+m+1)−(n−1))Π_(l=0) ^m (n+l)  =(m+2)Π_(l=0) ^m (n+l)        Σ_(n=1) ^k (Π_(l=0) ^m (n+l))  =(1/(m+2))Σ_(n=1) ^k (Π_(l=0) ^(m+1) (n+l)−Π_(l=0) ^(m+1) (n+l−1))  =(1/(m+2)) { (),() :}Π_(l=0) ^(m+1) (1+l)−Π_(l=0) ^(m+1) (1+l−1)                    +Π_(l=0) ^(m+1) (2+l)−Π_(l=0) ^(m+1) (2+l−1)                    +Π_(l=0) ^(m+1) (3+l)−Π_(l=0) ^(m+1) (3+l−1)                    +....+                    +Π_(l=0) ^(m+1) (k+l)−Π_(l=0) ^(m+1) (k+l−1) {: (),() }  =(1/(m+2)){Π_(l=0) ^(m+1) (k+l)−Π_(l=0) ^(m+1) (1+l−1)}  Π_(l=0) ^(m+1) (1+l−1)=0×1×...×(m+1)=0  ∴Σ_(n=1) ^k (Π_(l=0) ^m (n+l))=((Π_(l=0) ^(m+1) (k+l))/(m+2))

m+1l=0(n+l)m+1l=0(n+l1)=(n+m+1)ml=0(n+l)(n1)ml=0(n+l)=((n+m+1)(n1))ml=0(n+l)=(m+2)ml=0(n+l)kn=1(ml=0(n+l))=1m+2kn=1(m+1l=0(n+l)m+1l=0(n+l1))=1m+2{m+1l=0(1+l)m+1l=0(1+l1)+m+1l=0(2+l)m+1l=0(2+l1)+m+1l=0(3+l)m+1l=0(3+l1)+....++m+1l=0(k+l)m+1l=0(k+l1)}=1m+2{m+1l=0(k+l)m+1l=0(1+l1)}m+1l=0(1+l1)=0×1×...×(m+1)=0kn=1(ml=0(n+l))=m+1l=0(k+l)m+2

Commented by hassentimol last updated on 19/Jan/19

Oh... Thank you so much sir...  I am so grateful... God Bless You !!!

Oh...Thankyousomuchsir...Iamsograteful...GodBlessYou!!!

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