Mk ko hiểu đề bài bn ghi gì Nguyễn Như Quỳnh

Tính 

tinh tong 

S=0.2+2.4+4.6+.....+2n ( 2n+2 )

\(S=0.2+2.4+4.6+...+2n\left(2n+2\right)\)

\(6S=2.4.6+4.6.\left(8-2\right)+...+2n\left(2n+2\right)\left[\left(2n+4\right)-\left(2n-2\right)\right]\)

\(=2.4.6+4.6.8-2.4.6+...+2n\left(2n+2\right)\left(2n+4\right)-\left(2n-2\right).2n.\left(2n+2\right)\)

\(=2n\left(2n+2\right)\left(2n+4\right)\)

Suy ra \(S=\frac{2n\left(2n+2\right)\left(2n+4\right)}{6}\)

cái đề bài của bạn kiểu gì vậy hình như sai rồi

tính tổng cả dãy theo kiểu lớp 6

\(S=\frac{1}{2022}-\left(\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{2020.2022}\right)\)

\(A=\frac{5}{2.4}+\frac{5}{4.6}+...+\frac{5}{2020.2022}\)

\(A=\frac{5}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{2020.2022}\right)\)

\(A=\frac{5}{2}\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+...+\frac{2022-2020}{2020.2022}\right)\)

\(A=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2020}-\frac{1}{2022}\right)\)

\(A=\frac{5}{2}\left(\frac{1}{2}-\frac{1}{2022}\right)\)

\(S=\frac{1}{2022}-A=\frac{1}{2022}-\frac{5}{2}\left(\frac{1}{2}-\frac{1}{2022}\right)=-\frac{1262}{1011}\)

99/200

\(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+....+\frac{1}{98\cdot100}\)

\(=\frac{1}{2}\left(\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+.......+\frac{2}{98\cdot100}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+......+\frac{1}{98}-\frac{1}{100}\right)\)

\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{100}\right)=\frac{49}{200}\)

Đặt A = 1/2.4 + 1/4.6 + .... +1/98.100

\(2A=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\)

       \(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\)

        \(=\frac{1}{2}-\frac{1}{100}\)

         \(=\frac{49}{100}\)

\(\Rightarrow A=\frac{49}{100}\div2=\frac{49}{200}\)

Bài 1 :

\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)

Ta thấy :

\(3=2^2-1\)

\(15=4^2-1\)

\(35=6^2-1\)

.....

\(9999=100^2-1\)

\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)

\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)

\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)

nhanh len nhé mik đang cần gấp ai lam trước mik tích cho

Bài 6 :

\(C=1^2+2^2+...+100^2=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}=\dfrac{100.101.201}{6}=338350\)

Bài 9 :

\(S=1^2+2^2+3^2+...+99^2=\dfrac{99.\left(99+1\right)\left(2.99+1\right)}{6}=\dfrac{99.100.199}{6}=328350\)