Question

若{an}的首项为a1,公比为q的等比数列(q≠1),Sn是前n项和。求和S1Cn0-S2Cn1+S3cn2-S4Cn3……+(-1)^nS(n+1)

若{an}的首项为a1,公比为q的等比数列(q≠1),Sn是前n项和。
（1）由a1C20-a2C21+a3c22以及a1C30-a2C31+a3c32-a4C33的求和归纳出关于n的结论并证明
（2）求和S1Cn0-S2Cn1+S3cn2-S4Cn3……+(-1)^nS(n+1)Cnn

Answer 1

an=a1q^(n-1)
(1).a1C(2,0)-a2C(2,1)+a3C(2,2)
=a1C(2,0)-a1q*C(2,1)+a1q^2*C(2,2)
=a1[C(2,0)-qC(2,1)+q^2*C(2,2)
=a1(1-q)^2
a1C(3,0)-a2C(3,1)+a3C(3,2)-a4C(3,3)
=a1C(3,0)-a1q*C(3,1)+a1q^2*C(3,2)-a1q^3*C(3,3)
=a1(1-q)^3
一般结论：
a1C(n,0)-a2C(n,1)+a3C(n,2)-a4C(n,3)+.....+（-1)^(n-1)*C(n,n)=a1(1-q)^n
证明方法：展开右边，注意a1q^(n-1)=an即可。
(2).Sn=a1(1-q^n)/(1-q)=a1/(1-q)-a1q^n/(1-q)
S1C(n,0)-S2C(n,1)+S3C(n,2)-S4C(n,3)……+ (-1)^nS(n+1)C(n,n)
=C(n,0)a1/(1-q)-C(n,0)a1q/(1-q)-C(n,1)a1/(1-q)+C(n,1)a1q^2/(1-q)+
C(n,2)a1/(1-q)-C(n,2)a1q^3/(1-q)+......+ (-1)^nC(n,n)a1/(1-q)-(-1)^n*a1*q^nC(n,n)/(1-q)
=a1/(1-q)[C(n,0)-C(n,1)+C(n,2)-C(n,3)+.....(1-)^nC(n,)]+
[-C(n,0)a1q/(1-q)+C(n,1)a1q^2/(1-q)-C(n,2)a1q^3/(1-q)+......-(-1)^n*a1*q^nC(n,n)/(1-q)]
=a1/(1-q)[C(n,0)-C(n,1)+C(n,2)-C(n,3)+.....(1-)^nC(n,)]-a1*q/(1-q)[C(n,0)-C(n,1)q+C(n,2)q^2+......+(-1)^nq^nC(n,n)]
=a1q/1-q)*(1-1)^n-a1*q/(1-q)*(1-q)^n
=-a1q(1-q)^(n-1)