急由正整数组成的等比数列证明Sn乘S(n+2)小于S(n+1)^2
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急由正整数组成的等比数列证明Sn乘S(n+2)小于S(n+1)^2
是正整数组成的等比数列证明Sn乘S(n+2)小于(S(n+1))^2
是正整数组成的等比数列证明Sn乘S(n+2)小于(S(n+1))^2
![急由正整数组成的等比数列证明Sn乘S(n+2)小于S(n+1)^2](/uploads/image/z/20047996-28-6.jpg?t=%E6%80%A5%E7%94%B1%E6%AD%A3%E6%95%B4%E6%95%B0%E7%BB%84%E6%88%90%E7%9A%84%E7%AD%89%E6%AF%94%E6%95%B0%E5%88%97%E8%AF%81%E6%98%8ESn%E4%B9%98S%28n%2B2%29%E5%B0%8F%E4%BA%8ES%28n%2B1%29%5E2)
正整数组成的等比数列,a1>0 q>0
Sn=S(n+1)-a(n+1)
S(n+2)=S(n+1)+a(n+2)
Sn乘S(n+2)=S(n+1)^2-S(n+1)*a(n+1)*a(n+2)-a(n+1)*a(n+2)
Sn乘S(n+2)-S(n+1)^2
=-S(n+1)*a(n+1)*a(n+2)-a(n+1)*a(n+2)
Sn=S(n+1)-a(n+1)
S(n+2)=S(n+1)+a(n+2)
Sn乘S(n+2)=S(n+1)^2-S(n+1)*a(n+1)*a(n+2)-a(n+1)*a(n+2)
Sn乘S(n+2)-S(n+1)^2
=-S(n+1)*a(n+1)*a(n+2)-a(n+1)*a(n+2)
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