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If-C-r-stands-for-n-C-r-n-r-n-r-and-r-1-n-r-C-r-2-for-n-2-then-is-divisible-by-1-3-n-1-2-n-1-3-n-2n-1-4-n-2-1-

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Question Number 23884 by Tinkutara last updated on 09/Nov/17
If C_r stands for^n C_r = ((n!)/(r! n − r!)) and Σ_(r=1) ^n r.C_r ^2 = λ for n ≥ 2, then λ is divisible by (1) 3 (n − 1) (2) n + 1 (3) n (2n − 1) (4) n^2 + 1
IfCrstandsfornCr=n!r!n−r!and∑nr=1r.Cr2=λforn⩾2,thenλisdivisibleby(1)3(n−1)(2)n+1(3)n(2n−1)(4)n2+1
Answered by ajfour last updated on 09/Nov/17
n(1+x)^(n−1) (x+1)^n =(C_1 +2C_2 x+.. ...+nC_n x^(n−1) )(C_0 x^n +C_1 x^(n−1) +C_2 x^(n−2) +..) so coefficient of x^(n−1) in the expansion of n(1+x)^(2n−1) is =n^(2n−1) C_(n−1) =((n(2n−1)!)/(n!(n−1)!)) =Σ_(r=1) ^n rC_r ^( 2) =((n(2n−1)(2n−2)!)/(n(n−1)!(n−1)!)) =^(2n−2) C_(n−1) (2n−1) ⇒ λ is divisible by n and λ is divisible by (2n−1) . I cannot prove if it will be divisible by n(2n−1) or not !
n(1+x)n−1(x+1)n=(C1+2C2x+..Missing or unrecognized delimiter for \leftsocoefficientofxn−1intheexpansionofn(1+x)2n−1is=n2n−1Cn−1=n(2n−1)!n!(n−1)!=∑nr=1rCr2=n(2n−1)(2n−2)!n(n−1)!(n−1)!=2n−2Cn−1(2n−1)⇒λisdivisiblebynandλisdivisibleby(2n−1).Icannotproveifitwillbedivisiblebyn(2n−1)ornot!

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