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Simplify 02(n0)+12(n1)+22(n2)++n2(nn),
where n2.

Enter your answer in the form f(n)2g(n), where f(n) and g(n) are polynomials in n with integer coefficients.

 Aug 7, 2020
 #1
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Simplify 

 

02(n0)+12(n1)+22(n2)+23(n2)++(n1)2(nn1)+n2(nn),


where n2.

 

S=02(n0)+12(n1)+22(n2)+23(n2)++(n1)2(nn1)+n2(nn)S=nk=0k2(nk)S=nk=0kk(nk)|k=(k1)S=nk=0k(k1)(nk)|(k1)(nk)=(n1)(n1k1)S=nk=1k(n1)(n1k1)|(n1)=nS=nk=1kn(n1k1)S=nnk=1k(n1k1)

 

nk=1k(n1k1)= ?

(1+x)n1=(n10)x0+(n11)x1+(n12)x2++(n1n1)xn1|xx(1+x)n1=(n10)x1+(n11)x2+(n12)x3++(n1n1)xnx(1+x)n1=n1k=0(n1k)xk+1ddxx(1+x)n1=n1k=0(k+1)(n1k)xkx(n1)(1+x)n2+(1+x)n1=n1k=0(k+1)(n1k)xk|x=1(n1)(1+1)n2+(1+1)n1=n1k=0(k+1)(n1k)1k(n1)(2)n2+(2)n1=n1k=0(k+1)(n1k)(n1)2n2+2n1=nk=1k(n1k1)

 

S= ?

S=nnk=1k(n1k1)|nk=1k(n1k1)=(n1)2n2+2n1S=n((n1)2n2+2n1)S=n((n1)2n2+2n122)S=n((n1)2n2+2n22)S=n2n2(n1+2)S=n2n2(n+1)S=n(n+1)2n2|f(n)=n(n+1), g(n)=n2, n2

 

laugh

 Aug 10, 2020
edited by heureka  Aug 10, 2020
edited by heureka  Aug 10, 2020

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