Find the coefficient of x ^ 2 in the expansion of (1 + x) (1 + 2x) · (1 + NX)

Find the coefficient of x ^ 2 in the expansion of (1 + x) (1 + 2x) · (1 + NX)

Let u = 1 + 2 + 3 +... + n = n (n + 1) / 2 take any combination of 2 terms from 1 + X, 1 + 2x, 1 + 3x,. 1 + NX as the coefficient of x ^ 2, then the coefficient of x ^ 2 is [1 * (U-1) + 2 (U-2) + 3 (U-3) +... + n (U-N)] / 2 = [u ^ 2 - (1 * 1 + 2 * 2 +... + n * n)] / 2 = [n ^ 2 (n + 1) ^ 2 / 4 - n (n + 1) (2n + 1) /

1. In the expansion of (x + 1) (2x + 1). (NX + 1), what is the coefficient of the X term?
2. Given that the function f (x) (x belongs to R) satisfies f (1) = 1, and the derivative function f (x) of F (x) is less than 0.5, then the solution set of F (x) less than X / 2 + 0.5 is?

The coefficient of 1. X is 1 + 2 + + n = n(n + 1)/2.
2. Let g (x) = f (x) - (x + 1) / 2
Then G '(x) = f' (x) - 1 / 2 < 0
So g (x) decreases monotonically, and G (1) = f (1) - 1 = 0
So on (- ∞, 1), G (x) > G (1) = 0
On (1, + ∞), G (x) < g (1) = 0
That is, on (1, + ∞), f (x) < (x + 1) / 2
So the solution set of the original inequality is (1, + ∞)

1+x+2x+3x+…… +Nx = how much?