a(n)=a+(n-1)d,s(n)=na+n(n-1)d/2.證明.歸納法.n=1時(shí),s(1)=a(1)=a,結(jié)論成立.設(shè)n=k時(shí)結(jié)論成立,則s(k)=ka+k(k-1)d/2.n=k+1時(shí),s(k+1)=s(k)+a(k+1)=ka+k(k-1)d/2 + a+kd = (k+1)a + k(k-1)d/2 + 2kd/2 = (k+1)a + k[k-1+...麻煩不麻煩啊……我才小學(xué)耶……簡(jiǎn)單點(diǎn)好不……舉例子說明也行啊……小學(xué)就已經(jīng)開始學(xué)等差數(shù)列了嗎首尾相加除以2再乘以數(shù)字的個(gè)數(shù)唄這是3年級(jí)奧數(shù)等差數(shù)列的和=（首項(xiàng)+末項(xiàng)）×項(xiàng)數(shù)/2項(xiàng)數(shù)=（末項(xiàng)—首項(xiàng)）/公差+1公差=后項(xiàng)—首項(xiàng)等差數(shù)列的第n項(xiàng)=首項(xiàng)+（n—1）×公差首項(xiàng)=末項(xiàng)—公差×（項(xiàng)數(shù)—1）請(qǐng)寫出這五個(gè)公式的推導(dǎo)過程明天老師讓我們上臺(tái)講解不講解的會(huì)吃不了兜著走T_T哦 這樣簡(jiǎn)單不簡(jiǎn)單Sn=a1+a2+a3+……+an Sn=a1+(a1+d)+(a1+2d)+ ……+[a1+(n-1)d] Sn=an+(an-d)+(an-2d)+ ……+[an-(n-1)d] 以上兩式相加， 2Sn=(a1+an)+ (a1+an)+ …… + (a1+an) 則Sn==(a1+an)/2×n不簡(jiǎn)單……拜托用文字解釋……