ken_su
花朵
共回答了19个问题采纳率:84.2% 举报
(1)Sn=na1+(1/2)n(n-1)d
证明:1、当n=1时,s1=a1.成立
2、假设当n=k时,结论成立,即Sk=ka1+(1/2)k(k-1)d
则当n=k+1时,S(k+1)=Sk+a(k+1)
又因为ak=a1+(k-1)d
所以S(k+1)=ka1+(1/2)k(k-1)d+a1+kd
=(k+1)a1+kd((1/2)k-1/2+1)=(k+1)a1+(1/2)(k+1)kd
假设成立,所以Sn=na1+(1/2)n(n-1)d
(2)Sn=[a1(1-q的n次方)]除以(1-q)
证明:1、当n=1时,s1=a1.成立
2、假设当n=k时,结论成立,即Sk=[a1(1-q的k次方)]除以(1-q)
则当n=k+1时,S(k+1)=Sk+a(k+1)
又因为ak=a1q^(k-1)
所以S(k+1)=[a1(1-q^k)]/(1-q)+a1q^k=a1(1-q^k+q^k-q^(k+1))/(1-q)=a1(1-q^(k+1))/(1-q)
假设成立,所以Sn=[a1(1-q的n次方)]除以(1-q)
1年前
6