rr88
幼苗
共回答了17个问题采纳率:76.5% 举报
Sn=na1+n(n-1)d/2=na1+d/2*(n²-n)
那么Sp=pa1+d/2*(p²-p)
Sq=qa1+d/2*(q²-q)
所以Sp-Sq=a1*(p-q)+d/2*(p²-p-q²+q)
=a1*(p-q)+d/2*[(p+q)(p-q)-(p-q)]
=a1*(p-q)+d/2*(p-q)(q+q-1)
=(p-q)*[a1+d/2*(p+q-1)]
=0
而p≠q,即p-q≠0,所以a1+d/2*(p+q-1)=0
S(p+q)=(p+q)*a1+d/2*(p+q)(p+q-1)
=(p+q)*[a1+d/2*(p+q-1)]
=(p+q)*0
=0
1年前
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