Question

已知等差数列an=2n,令bn=an*x^n(x为实数）．求数列｛bn｝前n项和的公式．

Answer 1

n=2(nx^n)

S=2(x+2x^2+3x^3+……+nx^n)

xS=2[x^2+2x^3+3x^4+……+nx^(n+1)]

(1-x)S=2[x+x^2+x^3+……+x^n-nx^(n+1)]

=2(x-x^n)/(1-x)-2nx^(n+1)

=>S=2(x-x^n)/(1-x)^2-2nx^(n+1)/(1-x)