rút gọn biểu thức A=√(1+1/1^2+1/2^2)+√(1+1/2^2+1/3^2)+...+√(1+1/2006^2+1/2007^2) / là phân số

Đáp án: $A=2007-\dfrac1{2007}$

Giải thích các bước giải:

Xét hạng tử tổng quát:

$1+\dfrac1{n^2}+\dfrac1{(n+1)^2}$

$=(1+2\cdot \dfrac1n+\dfrac1{n^2})-\dfrac2n+\dfrac1{(n+1)^2}$

$=(1+\dfrac1n)^2-\dfrac2n+\dfrac1{(n+1)^2}$

$=(\dfrac{n+1}{n})^2-2\cdot \dfrac{n+1}n\cdot \dfrac1{n+1}+(\dfrac1{n+1})^2$

$=(\dfrac{n+1}n-\dfrac1{n+1})^2$

$=(1+\dfrac1n-\dfrac1{n+1})^2$

$\to \sqrt{1+\dfrac1{n^2}+\dfrac1{(n+1)^2}}=1+\dfrac1n-\dfrac1{n+1}$

Áp dụng ta có:

$A=(1+\dfrac11-\dfrac12)+(1+\dfrac12-\dfrac13)+...+(1+\dfrac1{2006}-\dfrac1{2007})$

$\to A=2006+(\dfrac11-\dfrac12+\dfrac12-\dfrac13+...+\dfrac1{2006}-\dfrac1{2007})$

$\to A=2006+(1-\dfrac1{2007})$

$\to A=2007-\dfrac1{2007}$