Question

额，第二小题怎么做， 不是只要答案，详细过程，我数学很差， 谢谢

Answer 1

Answer 2

a1+2a2+3a3+...+nan=n(n+1)(2n+1)
a1+2a2+3a3+...+(n-1)a(n-1)=(n-1)n(2n-1)

两式相减得：
nan=n(n+1)(2n+1)-(n-1)n(2n-1)
an=(n+1)(2n+1)-(n-1)(2n-1)=6n

Answer 3

选C
a1+2a2+3a3+......nan=n(n+1)(2n+1）
有a1+2a2+3a3+...nan+(n+1)a(n+1)=(n+1)(n+2)(2n+3)
拿下面的式子减去上面的式子得
(n+1)*a(n+1)=(n+1)(2n^2+7n+6-2n^2-n)=(n+1)(6n+6)
有a(n+1)=6(n+1)
可知an=6n

Answer 4

a1+2a2+3a3+....+nan=n(n+1)(2n+1).......(1)
a1+2a2+3a3+....+(n-1)an-1=n(n-1)(2n-1)....(2)
(1)-(2)得
nan=6n^2
an=6n
答案是CCCCCCCCCCCCC

Answer 5

希望能帮到你