Giải thích các bước giải:
Bài 128 :
a.Từ đề ta suy ra:
$A=\dfrac{2^2-1}{2^2}.\dfrac{3^2-1}{3^2}...\dfrac{n^2-1}{n^2}$
$\to A=\dfrac{(2-1)(2+1)}{2^2}.\dfrac{(3-1)(3+1)}{3^2}...\dfrac{(n-1)(n+1)}{n^2}$
$\to A=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}...\dfrac{(n-1)(n+1)}{n^2}$
$\to A=\dfrac{(1.2.3..(n-1)).(3.4.5..(n+1))}{(2.3.4..n)^2}$
$\to A=\dfrac{n+1}{3n}$
Câu 129 :
a.Ta có :
$\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{(n-1)n}$
$=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+..+\dfrac{n-(n-1)}{(n-1)n}$
$=1-\dfrac{1}2+\dfrac{1}2-\dfrac{1}3+..+\dfrac{1}{n-1}-\dfrac{1}n$
$=1-\dfrac{1}n$
Câu 130:
a.Ta có :
Giải thích các bước giải:
Ta có :
$A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+..+\dfrac{1}{(2n)^2}$
$\to A=\dfrac{1}{2^2}(1+\dfrac{1}{2^2}+..+\dfrac{1}{n^2})$
$\to A<\dfrac{1}{2^2}(1+\dfrac{1}{1.2}+..+\dfrac{1}{(n-1)n})$
$\to A<\dfrac{1}{2^2}(1+1-\dfrac{1}2+\dfrac{1}2-\dfrac{1}3+..+\dfrac{1}{n-1}-\dfrac{1}n)$
$\to A<\dfrac{1}{4}(2-\dfrac{1}n)$
$\to A<\dfrac{1}4.2=\dfrac{1}2$
