=\lim\limits_{t \to 0}\dfrac{1+nt+C_{n}^{2} t^2 + C_{n}^{3} t^3+...+ C_{n}^{n} t^n -nt-n +n -1}{t^2}
=\lim\limits_{t \to 0}\dfrac{C_{n}^{2} t^2 + C_{n}^{3} t^3+...+C_{n}^{n} t^n}{t^2}
=\lim\limits_{t \to 0} \left ( C_{n}^{2} +C_{n}^{3} t+...+C_{n}^{n} t^{n-2} \right )
=C_{n}^{2}=\dfrac{n(n-1)}{2}
Bài tập tương tự
1) \lim\limits_{x\to a}\dfrac{(x^n-a^n)-na^{n-1}(x-a)}{(x-a)^2}=\dfrac{n(n-1)a^{n-2}}{2}
2) \lim\limits_{x\to 0}\dfrac{(1+x)(1+2x)...(1+nx)-1}{x} =\dfrac{n(n+1)}{2}
3) \lim\limits_{x\to 0}\dfrac{(1+mx)^n - (1+nx)^m}{x^2}=\dfrac{mn(n-m)}{2}
đề hay quá thầy ơi. Câu 2 giải sao vậy thầy
Trả lờiXóaBài 2:
Trả lờiXóa\lim\limits_{x \to 0}\dfrac{(1+x)(1+2x)...(1+nx)-1}{x}
=\lim\limits_{x \to 0}\dfrac{1+(1+2+3+...+n)x+x^2.g(x)-1}{x}
=1+2+3+...+n=\dfrac{n(n+1)}{2}