Question

已知等差数列an集合的前3项和为6,前8项和为-4.求数列an集合的通项公式?设bn=(4-an)*q的n-1次方,求数

Answer 1

an列方程组计算得a1=3,d=-1

an=4-n

bn=nq^(n-1)

sn=1+2q+3q^2+4q^3+.+nq^(n-1)

kn=n+(n-1)q+(n-2)q^2+(n-3)q^3+.+2q^(n-2)+q^(n-1)

=(1-q^n)/(1-q)+(1-q^(n-1))/(1-q)+.+(1-q^2)/(1-q)+(1-q)/(1-q)

=(n-(q+q^2+.+q^n))/(1-q)

=(n(1-q)-q+q^(n+1))/(1-q)^2

sn+kn=(n+1)(1-q^n)/(1-q)

sn=(n+1)(1-q^n)/(1-q)-kn=(1-nq^n-q^n+nq^(n+1))/(1-q)^2