Do too much sum! Forget the basics?

Note that $n(n+2) + 1 = (n+1)^2$

$S_n = 1*3 + 2*4 + 3*5 + \ldots + n*(n+2)$

$\Rightarrow S_n = 1*3 + 1 - 1 + 2*4 + 1 - 1 + \ldots + n*(n+2) + 1 - 1$

$\Rightarrow S_n = 2^2 - 1 + 3^2 - 1 + \ldots + (n+1)^2 -1$

$\Rightarrow S_n = 2^2 + 3^2 + \ldots + (n+1)^2 - n$

$\Rightarrow S_n = 1^2 + 2^2 + 3^2 + \ldots + (n+1)^2 - (n+1)$

$\Rightarrow S_n = \dfrac{(n+1)(n+2)(2n + 3)}{6} - (n+1)$

Putting in $n = 30$

$S_{30} = 10416 - 31 = 10385$

Therefore answer $\boxed{385}$

just open the bracket, n(n+2)

$n^{2}$ + 2n

apply the summition

$\sum_{i=0}^n$ $n^{2}$ + $\sum_{i=0}^n$ 2n

$\sum_{i=0}^n$ $n^{2}$ + 2 $\sum_{i=0}^n$ n

$\frac{n(n+1)(2n+1)}{6} + 2 \frac{n(n+1)}{2}$

Now just put n=30

you will get 10385

so the ans is $\boxed{385}$

Lol ... just summation of n^2 + 2n from n=1 to 30 ...

$a \cdot 1^3+b \cdot 1^2+c \cdot 1+d=03$

$a \cdot 2^3+b \cdot 2^2+c \cdot 2+ d=11$

$a \cdot 3^3+b \cdot 3^2+c \cdot 3+d=26$

$a \cdot 4^3+b \cdot 4^2+c \cdot 4+d=50$

$a=\frac{1}{3}, b=\frac{3}{2}, c=\frac{7}{6}, d=0$

$a_n=\frac{1}{3}n^3+\frac{3}{2}n^2+\frac{7}{6}n$

$a_n=\frac{n}{6}(2n^2+9n+7)$

$a_{30}=10385$