Question

a1+2a2+3a3+4a4+...+nan=n

关于数列的问题
a1+2a2+3a3+4a4+……+nan=n(n+1)(n+2）,则an=
（注：a后的数字为项数）

Answer 1

a1+2a2+3a3+4a4+……+(n-1)a + nan=n(n+1)(n+2）,
a1+2a2+3a3+4a4+……+(n-1)a = (n-1)n(n+1)
两式相减
n an = n(n+1)(n+2) - (n-1)n(n+1) = 3n(n+1)
an = 3(n+1)