Question

等比数列的前N项和：求和（a-1)+(a^2-2)+.+(a^n-n)(2-3*5^(-1))+(4-3*5^(-2))

Answer 1

（a-1)+(a^2-2)+.+(a^n-n)

=(a+a^2.+a^n)-(1+2+.n)

=[a(1-a^n)/(1-a)]-[(1+n)n/2]

(2-3*5^(-1))+(4-3*5^(-2))+...(2n-3*5^(-n))

=2+4+...+2n-[(3*5^(-1)+(3*5^(-2)+...(3*5^(-n)]

=(1+n)n-[3/5((1-(1/5)^n)/1-1/5]

s1=1+2x+3x^2+...nx^(n-1)

s1*x=1x+2x^2+3x^3+nx^n

s1-s1x=(1+2x+3x^2+...nx^(n-1))-(1x+2x^2+3x^3+nx^n)

=1+x+x^2+x^3+...x^(n-1)-nx^n