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दिया गया श्रेणी है : `C_(0)+2.C_(1)+3.C_(2)+…+(n+1).C_(n)`
`=[(r-1)+1].””^(n)C_(r-1)=(r-1).””^(n)C_(r-1)+””^(n)C_(r-1)`
`=n.””^(n-1)C_(r-2)+””^(n)C_(r-1)” “[because (r-1).””^(n)C_(r-1)=n.””^(n-1)C_(r-2)]`
अब `” “C_(0)+2.C_(1)+3.C_(2)+…..+(n+1).C_(n)=underset(r=1)overset(n+1)sum t_(r)`
`=underset(r=1)overset(n+1)sumn.””^(n-1)C_(r-2)+underset(r=1)overset(n+1)sum””^(n)C_(r-1)=n underset(r=1)overset(n+1)sum””^(n-1)C_(r-2)+underset(r=1)overset(n+1)””^(n)C_(r-1)`
`=n[“”^(n-1)C_(0)+””^(n-1)C_(1)+…+””^(n-1)C_(n-1)]+(“”^(n)C_(0)+””^(n)C_(1)+…+””^(n)C_(n))`
`=n.2^(n-1)+2^(n)=2^(n-1)(n+2)`
समीकरण (C) तक (i) कि तरह लाइए,
`(C_(0)+C_(1)x+C_(2)x^(2)+……+C_(n)x^(n))`
`(C_(0)x^(n)+C_(1)x^(n-1)+……+C_(n))=(1+x)^(2n)” “…(C)`
दोनों तरफ से `x^(n)` का गुणांक बराबर करने पर,
`C_(0)””^(2)+C_(1)””^(2)+……+C_(n)””^(2)=””^(2n)C_(n)=(2n!)/(n!n1)`