第一問(wèn),求通項(xiàng)
因
a2 - a1 = 1
a3 - a2 = 2
a4 - a3 = 3
.
an - an-1 = n-1
將上式左右兩邊分別相加,得
an - a1 = 1+2+3+.+(n-1)
所以
an = (n-1)n/2 +a1 = (n-1)n/2
第二問(wèn),求和
因?yàn)?an = (n-1)n/2 = (1/2)n^2 - (1/2)n
所以
S = 1/2(1^2 + 2^2 + .+ n^2) - 1/2(1+2+3+.+n)
= (1/2)*[n(n+1)(2n+1)/6] - (1/2)*[n(n+1)/2]
= n(n^2 - 1)/6
= (n^3 - n)/6