C = \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+...+\left(1-\frac{1}{9900}\right)\)(99 CẶP)

     = \(\left(1+1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)\)(99 SỐ HẠNG 1)

     = \(1.99-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

     = \(99-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

     = \(99-\left(1-\frac{1}{100}\right)\)

     = \(99-\frac{99}{100}\)

     = \(\frac{9801}{100}\)

Vậy \(C=\frac{9801}{100}\)

Chúc bạn học tốt !!!!!

=(4950-1) cgia cho 2 +1=................................

1.

Áp dụng tính chất dãy tỉ số bằng nhau:
$\frac{a}{2}=\frac{b}{3}=\frac{c}{4}$

$=\frac{a}{2}=\frac{2b}{6}=\frac{3c}{12}=\frac{a+2b+3c}{2+6+12}=\frac{-20}{20}=-1$

$\Rightarrow a=2(-1)=-2; b=3(-1)=-3; c=4(-1)=-4$

2.

$S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{9900}$
$=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}$

$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+....+\frac{100-99}{99.100}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}$

$=1-\frac{1}{100}=\frac{99}{100}$

a) Ta có : \(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{c}{4}\)

\(\Rightarrow\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b+3c}{2+6+12}=\dfrac{-20}{20}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}a=\left(-1\right)\cdot2=-2\\b=\dfrac{\left(-1\right).6}{2}=-3\\c=\dfrac{\left(-1\right).12}{3}=-4\end{matrix}\right.\)

b) Ta có : \(S=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)

\(=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\).

Vậy : \(S=\dfrac{99}{100}.\)

a)\(\dfrac{a}{2}=\dfrac{2b}{6}=\dfrac{3c}{12}=\dfrac{a+2b+3c}{2+6+12}=-\dfrac{20}{20}=-1\)

\(\left\{{}\begin{matrix}\dfrac{a}{2}=-1\Leftrightarrow a=-2\\\dfrac{b}{3}=-1\Leftrightarrow b=-3\\\dfrac{c}{4}=-1\Leftrightarrow c=-4\end{matrix}\right.\)

b)\(S=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\\ =\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}\)

Ta có: 0 + 1 + 4 + 9 + ... + 5500 =1 + 4 + 9 + ... + 5500 

Tổng trên có số số hạng là: 

Số số hạng của dãy số 0+1+4+9+...+5500 là

(5500-1):3+1=1834(số)

Tổng là

(5500+1) x 1834:2=5044417

A ] { 5500 - 1 } : 3 + 1 = 1834 số

{ 5500 + 1 } x 1834 : 2 = 5044417

B ] ? 

\(\left(1-\frac{2}{6}\right)\left(1-\frac{2}{12}\right)...\left(1-\frac{2}{9900}\right)\)

\(=\frac{4}{6}.\frac{10}{12}...\frac{9898}{9900}\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}...\frac{98.101}{99.100}\)

\(=\frac{1.2...98}{3.4...100}.\frac{4.5...101}{2.3...99}\)

\(=\frac{2}{99.100}.\frac{100.101}{2.3}\)

\(=\frac{101}{99.3}\)

\(=\frac{101}{297}\)

đáp số:\(\frac{101}{297}\)

ai k mk mk sẽ k lại ^-^

là 101/247

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