$S=\dfrac{−nC1}{2.3}+\dfrac{2.nC2}{3.4}−\dfrac{3.nC3}{4.5}+...+\dfrac{(−1)n.n.nCn}{(n+1)(n+2)}$
$A=(1-x)^n=C_n^0-C_n^1x+C_n^2x^2-C_n^3x^3+...+(-1)^nC_n^nx^n$
$B=\int_{0}^{x}Adx=-\dfrac{(1-x)^{n+1}}{n+1}\big|_0^x=-\dfrac{(1-x)^{n+1}}{n+1}+\dfrac{1}{n+1}$
$=C_n^0x-\dfrac{C_n^1x^2}{2}+...+\dfrac{(-1)^nC_n^nx^{n+1}}{n+1}$
$C=\int_{0}^{x}Bdx=\dfrac{(1-x)^{n+2}}{(n+1)(n+2)}+\dfrac{x}{n+1}\big|_0^x$
$=\dfrac{(1-x)^{n+2}}{(n+1)(n+2)}-\dfrac{1}{(n+1)(n+2)}+\dfrac{x}{n+1}$
$=\dfrac{C_n^0x^2}{2}-\dfrac{C_n^1x^3}{2.3}+...+\dfrac{(-1)^nC_n^nx^{n+2}}{(n+1)(n+2)}$
$D=(\dfrac{C}{x^2})'=\dfrac{1}{(n+1)(n+2)}[(\dfrac{-(n+2)(1-x)^{n+1}x^2-2x(1-x)^{n+2}}{x^4})+2x^{-3}]+\dfrac{lnx}{n+1}$
$D=-\dfrac{C_n^1}{2.3}+\dfrac{2C_n^2x}{3.4}+...+\dfrac{(-1)^nnC_n^nx^{n-1}}{(n+1)(n+2)}$
Cho $x=1\to S=\dfrac{2}{(n+1)(n+2)}$ ycbt
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn