an=6+3+5+9...+(n²-3n+5 )=6+【1²+2²...+(n-1)²】-3【1+2+...(n-1)】+5(n-1)?

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/28 02:18:05
an=6+3+5+9...+(n²-3n+5 )=6+【1²+2²...+(n-1)²】-3【1+2+...(n-1)】+5(n-1)?

an=6+3+5+9...+(n²-3n+5 )=6+【1²+2²...+(n-1)²】-3【1+2+...(n-1)】+5(n-1)?
an=6+3+5+9...+(n²-3n+5 )=6+【1²+2²...+(n-1)²】-3【1+2+...(n-1)】+5(n-1)?

an=6+3+5+9...+(n²-3n+5 )=6+【1²+2²...+(n-1)²】-3【1+2+...(n-1)】+5(n-1)?
an=6+3+5+9+...+(n²-3n+5 )=6+(2²-3*2+5)+(3²-3*3+5)+...+(n²-3n+5 )=6+(2²+3²...+n²)-3(2+3+...+n)+5(n-1)