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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
∏m+1l=0(n+l)−∏m+1l=0(n+l−1)=(n+m+1)∏ml=0(n+l)−(n−1)∏ml=0(n+l)=((n+m+1)−(n−1))∏ml=0(n+l)=(m+2)∏ml=0(n+l)∑kn=1(∏ml=0(n+l))=1m+2∑kn=1(∏m+1l=0(n+l)−∏m+1l=0(n+l−1))=1m+2{∏m+1l=0(1+l)−∏m+1l=0(1+l−1)+∏m+1l=0(2+l)−∏m+1l=0(2+l−1)+∏m+1l=0(3+l)−∏m+1l=0(3+l−1)+....++∏m+1l=0(k+l)−∏m+1l=0(k+l−1)}=1m+2{∏m+1l=0(k+l)−∏m+1l=0(1+l−1)}∏m+1l=0(1+l−1)=0×1×...×(m+1)=0∴∑kn=1(∏ml=0(n+l))=∏m+1l=0(k+l)m+2
Commented by hassentimol last updated on 19/Jan/19
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