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Question Number 56137 by gunawan last updated on 11/Mar/19

If (1+2x+x^2 )^n = Σ_(r=0) ^(2n) a_r x^r , then a_r =

If(1+2x+x2)n=∑2nr=0arxr,thenar=

Commented by maxmathsup by imad last updated on 11/Mar/19

(1+2x+x^2 )^n =(x+1)^(2n) =Σ_(r=0) ^(2n) C_(2n) ^r x^r ⇒a_r =C_(2n) ^r =(((2n)!)/(r!(2n−r)!)) and r∈[[0,2n]].

(1+2x+x2)n=(x+1)2n=∑r=02nC2nrxr⇒ar=C2nr=(2n)!r!(2n−r)!andr∈[[0,2n]].

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Mar/19

(1+2x+x^2 )^n (1+x)^(2n) =a_0 x^0 +a_1 x^1 +a_2 x^2 +...+a_(2n) x^(2n) put x=1 2^(2n) =Σ_(r=0) ^(2n) a_r r+1 the term 2nc_r x^(2n) so a_r =2nc_r =((2n!)/(r!(2n−r)!))

(1+2x+x2)n(1+x)2n=a0x0+a1x1+a2x2+...+a2nx2nputx=122n=∑2nr=0arr+1theterm2ncrx2nsoar=2ncr=2n!r!(2n−r)!

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