Giải thích các bước giải:

Ta có:

$A=\dfrac54+\dfrac{10}9+\dfrac{17}{16}+...+\dfrac{2501}{2500}$

$\to A=\dfrac{4+1}4+\dfrac{9+1}9+\dfrac{16+1}{16}+...+\dfrac{2500+1}{2500}$

$\to A=(1+\dfrac{1}4)+(1+\dfrac{1}9)+(1+\dfrac{1}{16})+...+(1+\dfrac{1}{2500})$

$\to A=(1+\dfrac{1}{2^2})+(1+\dfrac{1}{3^2})+(1+\dfrac{1}{4^2})+...+(1+\dfrac{1}{50^2})$

$\to A=49+(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2})$

Mà $B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}$

$\to B=\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{50.50}$

$\to B<\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}$

$\to B<\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+...+\dfrac{50-49}{49.50}$

$\to B<\dfrac11-\dfrac12+\dfrac12-\dfrac13+...+\dfrac1{49}-\dfrac1{50}$

$\to B<1-\dfrac1{50}<1$

Lại có: $B>0\to 0<B<1\to B$ không là số nguyên

Mà $A=49+B\to A$ không là số nguyên