What is the coefficient of x^(n–r) in the expansion of (x + 1)(x + 2)(x + 3) . . . (x + n) ?

Expansion of (x+1)(x+2)(x+3)(x+4)............(x+n) =

= (x^n) + [{1+2+3+......+n}*{x^(n-1)}] + [{1*2+1*3+1*4+ --+2*3+2*4+ --+(n-1)*(n)}*{x^(n-2)}] + --

In general it is: (x^n) + {(n)(n+1)/2}*{x^(n-1)} + {Sum of products of two nos, taken at a time; there will be nC2 factors}*{n^(n-2)} + {Sum of products of 3 nos. taken at a time, there will be nC3 factors}*{n^(n-3)} + ..

Hence, coefficient of x^(n-r) is:

Sum of coefficient of 'r' nos. taken at a time; there will be nCr such factors.

Ex. Let n = 5 and r = 3; ===> Coefficient of of x^(5-3)

So there has to be 5C3 factors of triplets = 10 triplets

They are {(123), (124), (125), (134), (135), (145), (234), (235), (245), (345)}

Sum of the products of each triplet =

= 6 + 8 + 10 + 12 + 15 + 20 + 24 + 30 + 40 + 60 = 225

its 1, 2 and 3

P(x) = (x + 1)(x + 2)(x + 3) . . . (x + n)

Using relations between roots ( zero) and coefficients ( Viete or Vieta )

x^n has coefficient 1

x^(n-1) has coefficient 1+2 + 3+ ....+ n = n(n+1)/2

x^(n-2) has coefficient 1*2+ 1*3+ .. 1*n + 2*3+ .. 2*n +..... (n-1)*n

it's sum of any double products x_i*x_j where i < j for i between 1,n-1 and j between 2 and n

x^(n-r) is sum of products x_i_1 * x_i_2* .... *x_i_r where i_1< i_2<i_3 < ....i_r for

i_1 from1 to n-r

i_2 from 2 to n-r+1

,

i_r from r to n

last coefficient is n! where x has degree 0

Sono stata scambiata according to una mazza da baseball perché sono uniqueness (^.^) P.s perché non riesco a vedere le tue risposte nelle domande :-( Alle domande alle quali tu rispondi io non le vedo :-( A le domande in generale..dopo posto una domanda e se rispondi vedo se anche alle mie non comparì...oppure si... Certo (^.^)

jai ho