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If-P-n-denotes-the-product-of-the-binomial-coefficients-in-the-expansion-of-1-x-n-then-P-n-1-P-n-equals-




Question Number 52600 by Tip Top last updated on 10/Jan/19
If P_n  denotes the product of the binomial   coefficients in the expansion of (1+x)^n ,  then (P_(n+1) /P_n ) equals
IfPndenotestheproductofthebinomialcoefficientsintheexpansionof(1+x)n,thenPn+1Pnequals
Commented by Abdo msup. last updated on 10/Jan/19
we have (x+1)^n  =Σ_(k=0) ^n  C_n ^k  x^k  ⇒  P_n =Π_(k=0) ^n  C_n ^k   =C_n ^o  .C_n ^1  .....C_n ^n  ⇒  P_(n+1)  =C_(n+1) ^0  .C_(n+1) ^1   ...C_(n+1) ^(n+1)  ⇒  (P_(n+1) /P_n ) =((Π_(k=0) ^n  C_n ^k )/(Π_(k=0) ^(n+1)  C_(n+1) ^k )) =((Π_(k=0) ^n  C_n ^k )/(Π_(k=0) ^n   (((n+1)!)/(k!(n+1−k)!))))  =((Π_(k=0) ^n  C_n ^k )/(Π_(k=0) ^n   ((n+1)/((n+1−k)))  Π_(k=0) ^n  ((n!)/(k!(n−k)!))))  =Π_(k=0) ^n   (((n+1−k))/(n+1)) = (((n+1)!)/((n+1)^(n+1) )) =(((n+1)n!)/((n+1)(n+1)^n ))  =((n!)/((n+1)^n )) .
wehave(x+1)n=k=0nCnkxkPn=k=0nCnk=Cno.Cn1..CnnPn+1=Cn+10.Cn+11Cn+1n+1Pn+1Pn=k=0nCnkk=0n+1Cn+1k=k=0nCnkk=0n(n+1)!k!(n+1k)!=k=0nCnkk=0nn+1(n+1k)k=0nn!k!(nk)!=k=0n(n+1k)n+1=(n+1)!(n+1)n+1=(n+1)n!(n+1)(n+1)n=n!(n+1)n.
Answered by tanmay.chaudhury50@gmail.com last updated on 10/Jan/19
p_n =nc_0 ×nc_1 ×nc_2 ...nc_n   ←product of nterms  p_(n+1) =n+1c_0 ×n+1c_1 ×n+1c_2 ..×n+1c_(n+1) ←product of (n+1)terms  now ratio of r+1 th term    =((n+1c_r )/(nc_r ))  =((((n+1)!)/(r!(n+1−r)!))/((n!)/(r!(n−r)!)))=(((n+1)!)/(n!))×(((n−r)!)/((n+1−r)!))  =((n+1)/(n+1−r))  so (p_(n+1) /p_n ).=(/)  =((n+1)/(n+1−0))×((n+1)/(n+1−1))×((n+1)/(n+1−2))...  =(((n+1)^n )/((n+1)!))
pn=nc0×nc1×nc2ncnproductofntermspn+1=n+1c0×n+1c1×n+1c2..×n+1cn+1productof(n+1)termsnowratioofr+1thterm=n+1crncr=(n+1)!r!(n+1r)!n!r!(nr)!=(n+1)!n!×(nr)!(n+1r)!=n+1n+1rsopn+1pn.==n+1n+10×n+1n+11×n+1n+12=(n+1)n(n+1)!

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