1,4,10,20,35.规律式怎么求?
5个回答
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a3-a1 = 10-1 = 9 = 3²
a4-a2 = 20-4 = 16 = 4²
a5-a3 = 35-10 = 25 = 5²
.
an-1 -an-3 = (n-1)²
an - an-2 = n²
上式相加：an +an-1 -a2 - a1=3²+4²+5²+...+n²
an +an-1 =1+4+3²+4²+5²+...+n² =1²+2²+3²+4²+5²+...+n² = n(n+1)(2n+1)/6.①
a1=1
a2=1+(1+2)
a3=1+(1+2)+(1+2+3)
.
an-1 = 1+(1+2)+(1+2+3)+...+(1+2+3+...+n-1)
an = 1+(1+2)+(1+2+3)+...+(1+2+3+...+n-1) + (1+2+3+...+n)
得：an - an-1 = 1+2+3+...+n =n(n+1)/2 .②
①+②得：2an = n(n+1)(2n+1)/6 + n(n+1)/2
an = n(n+1)(n+2)/6
1²+2²+3²+4²+5²+...+n² = n(n+1)(2n+1)/6 也可以推导出来
