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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
$$\mathrm{If}\:{C}_{{r}} \:\mathrm{stands}\:\mathrm{for}\:^{{n}} {C}_{{r}} \:=\:\frac{{n}!}{{r}!\:{n}\:−\:{r}!}\:\mathrm{and} \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{r}.{C}_{{r}} ^{\mathrm{2}} \:=\:\lambda\:\mathrm{for}\:{n}\:\geqslant\:\mathrm{2},\:\mathrm{then}\:\lambda\:\mathrm{is}\:\mathrm{divisible} \\ $$$$\mathrm{by} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{3}\:\left({n}\:−\:\mathrm{1}\right) \\ $$$$\left(\mathrm{2}\right)\:{n}\:+\:\mathrm{1} \\ $$$$\left(\mathrm{3}\right)\:{n}\:\left(\mathrm{2}{n}\:−\:\mathrm{1}\right) \\ $$$$\left(\mathrm{4}\right)\:{n}^{\mathrm{2}} \:+\:\mathrm{1} \\ $$
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 !
$${n}\left(\mathrm{1}+{x}\right)^{{n}−\mathrm{1}} \left({x}+\mathrm{1}\right)^{{n}} =\left({C}_{\mathrm{1}} +\mathrm{2}{C}_{\mathrm{2}} {x}+..\right. \\ $$$$\left….+{nC}_{{n}} {x}^{{n}−\mathrm{1}} \right)\left({C}_{\mathrm{0}} {x}^{{n}} +{C}_{\mathrm{1}} {x}^{{n}−\mathrm{1}} +{C}_{\mathrm{2}} {x}^{{n}−\mathrm{2}} +..\right) \\ $$$${so}\:{coefficient}\:{of}\:{x}^{{n}−\mathrm{1}} \:{in}\:{the} \\ $$$${expansion}\:{of}\:{n}\left(\mathrm{1}+{x}\right)^{\mathrm{2}{n}−\mathrm{1}} \:\:{is} \\ $$$$\:\:={n}^{\mathrm{2}{n}−\mathrm{1}} {C}_{{n}−\mathrm{1}} \:=\frac{{n}\left(\mathrm{2}{n}−\mathrm{1}\right)!}{{n}!\left({n}−\mathrm{1}\right)!}\:=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{rC}_{{r}} ^{\:\mathrm{2}} \\ $$$$\:\:\:=\frac{{n}\left(\mathrm{2}{n}−\mathrm{1}\right)\left(\mathrm{2}{n}−\mathrm{2}\right)!}{{n}\left({n}−\mathrm{1}\right)!\left({n}−\mathrm{1}\right)!} \\ $$$$\:\:=^{\mathrm{2}{n}−\mathrm{2}} {C}_{{n}−\mathrm{1}} \left(\mathrm{2}{n}−\mathrm{1}\right) \\ $$$$\Rightarrow\:\:\lambda\:{is}\:{divisible}\:{by}\:{n}\:{and} \\ $$$$\lambda\:{is}\:{divisible}\:{by}\:\:\:\:\left(\mathrm{2}{n}−\mathrm{1}\right)\:. \\ $$$${I}\:{cannot}\:{prove}\:{if}\:{it}\:{will}\:{be} \\ $$$${divisible}\:{by}\:{n}\left(\mathrm{2}{n}−\mathrm{1}\right)\:{or}\:{not}\:! \\ $$

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