Toán 8 Tính

H

hotien217

f(x)=ax2+bxf_{(x)}=ax^2+bx;
f(x1)=ax22ax+a+bxbf_{(x-1)}=ax^2-2ax+a+bx-b
f(x)f(x1)=2axa+bf_{(x)}-f_{(x-1)}=2ax-a+b
Ta có: f(x)f_{(x)}-f(x1)f_{(x-1)}=xx \Rightarrow 2axa+b=x2ax-a+b=x
\Rightarrow a=b=12a=b=\dfrac{1}{2}
Thay vào f(x)f_{(x)} ta có:
f(x)=12x2+12xf_{(x)}=\dfrac{1}{2}x^2+\dfrac{1}{2}x
=12x(x+1)=x(x+1)2\dfrac{1}{2}x(x+1)=\dfrac{x(x+1)}{2}
ta sẽ suy ra được công thức tính:
1+2+3+...+n=x(x+1)21+2+3+...+n=\dfrac{x(x+1)}{2}
 
Last edited by a moderator:
H

huynhbachkhoa23

f(x)=ax2+bxf_{(x)}=ax^2+bx;
f(x1)=ax22ax+a+bxbf_{(x-1)}=ax^2-2ax+a+bx-b
f(x)f(x1)=2axa+bf_{(x)}-f_{(x-1)}=2ax-a+b
Ta có: f(x)f_{(x)}-f(x1)f_{(x-1)}=xx \Rightarrow 2axa+b=x2ax-a+b=x
\Rightarrow a=b=12a=b=\dfrac{1}{2}
Thay vào f(x)f_{(x)} ta có:
f(x)=12x2+12xf_{(x)}=\dfrac{1}{2}x^2+\dfrac{1}{2}x
=12x(x+1)=x(x+1)2\dfrac{1}{2}x(x+1)=\dfrac{x(x+1)}{2}
ta sẽ suy ra được công thức tính:
1+2+3+...+n=x(x+1)21+2+3+...+n=\dfrac{x(x+1)}{2}

Rõ hơn một chút:

1+2+3+...+n=f(1)f(0)+f(2)f(1)+f(3)f(2)+...+f(n)f(n1)=f(n)f(0)=f(n)=n(n+1)21+2+3+...+n=f(1)-f(0)+f(2)-f(1)+f(3)-f(2)+...+f(n)-f(n-1)=f(n)-f(0)=f(n)=\dfrac{n(n+1)}{2}
 
Top Bottom