Kí hiệu n!=1.2.3…..n với n là số nguyên dương . So sánh các phân số sau
M=$\frac{M=1.1!+2.2!+3.3!+…..+100.100!}{1.199+2.197+3.195+…..+100.1}$ , N=$\frac{99!}{33}$
Kí hiệu n!=1.2.3…..n với n là số nguyên dương . So sánh các phân số sau
M=$\frac{M=1.1!+2.2!+3.3!+…..+100.100!}{1.199+2.197+3.195+…..+100.1}$ , N=$\frac{99!}{33}$
Đáp án: $M<N$
Giải thích các bước giải:
Ta có:
$A=1\cdots 199+2\cdots 197+3\cdots 195+\cdots \cdots \cdots +100\cdots 1$
$\to A=1\cdots (199-0)+2\cdots (198-1)+3\cdots (197-2)+\cdots \cdots \cdots +100\cdots (100-99)$
$\to A=1\cdots 199-1\cdots 0+2\cdots 198-2\cdots 1+3\cdots 197-3\cdots 2+\cdots \cdots \cdots +100\cdots 100-100\cdots 99$
$\to A=(1\cdots 199+2\cdots 198+\cdots \cdots \cdots +100\cdots 100)-(0\cdots 1+1\cdots 2+\cdots \cdots \cdots +99\cdots 100)$
$\to A=(1\cdots (200-1)+2\cdots (200-2)+\cdots \cdots \cdots +100\cdots (200-100))-(0\cdots 1+1\cdots 2+\cdots \cdots \cdots +99\cdots 100)$
$\to A=(1\cdots 200-1^2+2\cdots 200-2^2+\cdots \cdots \cdots +100\cdots 200-100^2)-(0\cdots 1+1\cdots 2+\cdots \cdots \cdots +99\cdots 100)$
$\to A=(1\cdots 200+2\cdots 200+\cdots \cdots \cdots +100\cdots 200)-(1^2+2^2+\cdots \cdots \cdots +100^2)-(0\cdots 1+1\cdots 2+\cdots \cdots \cdots +99\cdots 100)$
$\to A=200(1+2+\cdots \cdots +100)-(1\cdots (1+0)+2(1+1)+\cdots \cdots \cdots +100\cdots (99+1))-(0\cdots 1+1\cdots 2+\cdots \cdots \cdots +99\cdots 100)$
$\to A=200(1+2+\cdots \cdots +100)-(1\cdots 1+1\cdots 0+2\cdots 1+2\cdots 1+\cdots \cdots \cdots +100\cdots 99+100\cdots 1)-(0\cdots 1+1\cdots 2+\cdots \cdots \cdots +99\cdots 100)$
$\to A=200(1+2+\cdots \cdots +100)-((1\cdots 1+2\cdots 1+\cdots \cdots \cdots +100\cdots 1)+(0\cdots 1+1\cdots 2+\cdots \cdots +99\cdots 100))-(0\cdots 1+1\cdots 2+\cdots \cdots \cdots +99\cdots 100)$
$\to A=200(1+2+\cdots \cdots +100)-((1+2+\cdots \cdots \cdots +100)+(0\cdots 1+1\cdots 2+\cdots \cdots +99\cdots 100))-(0\cdots 1+1\cdots 2+\cdots \cdots \cdots +99\cdots 100)$
$\to A=200(1+2+\cdots \cdots +100)-(1+2+\cdots \cdots \cdots +100)-(0\cdots 1+1\cdots 2+\cdots \cdots +99\cdots 100)-(0\cdots 1+1\cdots 2+\cdots \cdots \cdots +99\cdots 100)$
$\to A=200(1+2+\cdots \cdots +100)-(1+2+\cdots \cdots \cdots +100)-2(0\cdots 1+1\cdots 2+\cdots \cdots +99\cdots 100)$
$\to A=200(1+2+\cdots \cdots +100)-(1+2+\cdots \cdots \cdots +100)-2(1\cdots 2+\cdots \cdots +99\cdots 100)$
$\to A=(200-1)(1+2+\cdots \cdots +100)-2(1\cdots 2+\cdots \cdots +99\cdots 100)$
$\to A=199\cdot \dfrac{100(100+1)}{2}-2(1\cdots 2+\cdots \cdots +99\cdots 100)$
$\to A=199\cdot 50 \cdot 101-2(1\cdots 2+\cdots \cdots +99\cdots 100)$
Ta có $B=1\cdots 2+\cdots \cdots +99\cdots 100$
$\to B=1\cdots 2+2\cdots 3+\cdots \cdots \cdots +99\cdots 100$
$\to 3B=3(1\cdots 2+2\cdots 3+\cdots \cdots +99\cdots 100)$
$\to 3B=3\cdots 1\cdots 2++3\cdots 2\cdots 3+\cdots \cdots \cdots +3\cdots 99\cdots 100$
$\to 3B=(3-0)\cdots 1\cdots 2++(4-1)\cdots 2\cdots 3+\cdots \cdots \cdots +(101-98)\cdots 99\cdots 100$
$\to 3B=-0\cdots 1\cdots 2+1\cdots 2\cdots 3-1\cdots 2\cdots 3+2\cdots 3\cdots 4+\cdots \cdots \cdots \cdots -98\cdots 99\cdots 100+99\cdots 100\cdots 101$
$\to 3B=99\cdots 100\cdots 101$
$\to B=33\cdots 100\cdots 101$
$\to A=199\cdot 50 \cdot 101-33\cdots 100\cdots 101$
$\to A=101\cdot 50\cdot(199- 33\cdot 2)$
$\to A=101\cdot 50\cdot(199- 66)$
$\to A=101\cdot 50\cdot 133$
Ta có:
$C=1\cdots 1!+2\cdots 2!+3\cdots 3!+\cdots \cdots \cdots +100\cdots 100!$
$\to C=(2-1)\cdots 1!+(3-1)\cdots 2!+(4-1)\cdots 3!+\cdots \cdots \cdots +(101-1)\cdots 100!$
$\to C=2!-1!+3!-2!+4!-3!+\cdots \cdots \cdots +101!-100!$
$\to C=-1!+2!-2!+3!-3!+4!+\cdots \cdots \cdots +100!-100!+101!$
$\to C=-1!+101!$
$\to C=-1+101!$
$\to M=\dfrac{-1+101!}{101\cdot 50\cdot 133}$
$\to M<\dfrac{101!}{101\cdot 50\cdot 133}$
$\to M<\dfrac{100!}{ 50\cdot 133}$
Mà $50\cdot 133>33\cdot 100$
$\to M<\dfrac{100!}{ 33\cdot 100}$
$\to M<\dfrac{99!}{ 33}$
$\to M<N$