Помогите Срочно плиз (a+2a+3a+...+na)/(n^2-2n-3)-3a/(2(n-3))

a + 2a + 3a + 4a +  ... + na = a·(1+2+3+4+ ... + n)=a·(1+n)·n/2 n²-2n-3=0 D=(-2)²-4·(-3)=4+12=16 n₁=(2-4)/2=-1    или       n₂=(2+4)/2=3 n²-2n-3=(n+1)(n-3) [latex] \frac{a+2a+3a+...+na}{n^2-2n-3}- \frac{3a}{2(n-3)} = \\ \\ =\frac{a+2a+3a+...+na}{n^2-2n-3}- \frac{3a}{2(n-3)} =\frac{a\cdot \frac{(1+n)\cdot n}{2} }{(n+1)(n-3)}- \frac{3a}{2(n-3)} = \\ \\ =\frac{a\cdot n}{2\cdot(n-3)}- \frac{3a}{2(n-3)} = \\ \\ = \frac{an-3a}{2\cdot(n-3)}= \frac{a\cdot(n-3)}{2\cdot(n-3)}= \frac{a}{2} [/latex]