Câu hỏi của Thi Ngọc Hà Phương

Question

<p style="border:0px;color:rgba(0,0,0,0.87);">a) Tính tổng: <img alt="M=1+\frac{1}{5}+\frac{3}{35}+\ldots+\frac{3}{9603}+\frac{3}{9999}" width="318" height="41" class="lazy lightbox loaded" style="bor...

Yen Nhi · Accepted Answer

`Answer:`

a) $M=1+\frac{1}{5}+\frac{3}{35}+...+\frac{3}{9603}+\frac{3}{9999}$

$=\frac{3}{3}+\frac{3}{15}+\frac{3}{35}+...+\frac{3}{9603}+\frac{3}{9999}$

$=\frac{3}{2}.\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{9603}+\frac{2}{9999}\right)$

$=\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}+\frac{2}{99.101}\right)$

$=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{99}-\frac{1}{101}\right)$

$=\frac{3}{2}.\left(1-\frac{1}{101}\right)$

$=\frac{3}{2}.\frac{100}{101}$

$=\frac{150}{101}$

b) $S=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{\left(2n\right)^2}$

$=\frac{1}{2^2.2^2}+\frac{1}{2^2.3^2}+\frac{1}{2^2.4^2}+...+\frac{1}{2^2.n^2}$

$=\frac{1}{4.2^2}+\frac{1}{4.3^2}+\frac{1}{4.4^2}+...+\frac{1}{4n^2}$

$=\frac{1}{4}.\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\right)$

$=\frac{1}{4}.\left(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{n.n}\right)$

$\Rightarrow S< \frac{1}{4}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\right)$

$\Rightarrow S< \frac{1}{4}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)$

$\Rightarrow S< \frac{1}{4}.\left(1-\frac{1}{n}\right)$

$\Rightarrow S< \frac{1}{4}-\frac{1}{4n}< \frac{1}{4}$

$\Rightarrow S< \frac{1}{4}$

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