Tính tổng :\(S=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+....+\frac{1}{50}.\left(1+2+3+4+....+50\right)\)
Những câu hỏi liên quan
Tính tổng S=\(\sqrt{1+\left(1+\frac{1}{3}\right)^2}+\sqrt{1+\left(\frac{1}{2}+\frac{1}{4}\right)^2}+\sqrt{1+\left(\frac{1}{3}+\frac{1}{5}\right)^2}+...+\sqrt{1+\left(\frac{1}{48}+\frac{1}{50}\right)^2}\)
giúp mình với
- Tính nhanh tổng sau:
\(\left(\frac{1}{2}-1\right):\left(\frac{1}{3}-1\right):\left(\frac{1}{4}-1\right):\left(\frac{1}{5}-1\right)...:\left(\frac{1}{50}-1\right)\)
tìm x:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{x.\left(x+1\right):2}=\frac{1991}{1993}\)
Tính tổng :
\(\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{3^2}\right)+\left(1-\frac{1}{4^2}\right)+....+\left(1-\frac{1}{50^2}\right)\)
Câu 1:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=\frac{1991}{1993}.\)
\(\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{x.\left(x+1\right):2}=\frac{1991}{1993}\)
\(\frac{1}{2.3}.2+\frac{1}{3.4}.2+\frac{1}{4.5}.2+...+\frac{1}{x.\left(x+1\right)}.2=\frac{1991}{1993}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{1991}{1993}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1991}{3986}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1991}{3986}\)
...
e tự tính nốt nha
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\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{1991}{1993}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1991}{1993}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1991}{1993}\div2\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1991}{3986}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1991}{3986}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1991}{3986}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1991}{3986}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{1993}\)
\(\Leftrightarrow x+1=1993\)
\(\Leftrightarrow x=1993-1\)
\(\Leftrightarrow x=1992\)
Vậy x = 1992
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Tính:
\(\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right).\left(\frac{1}{5^2}-1\right)...\left(\frac{1}{50^2}-1\right)\)
\(\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right).\left(\frac{1}{5^2}-1\right)...\left(\frac{1}{50^2}-1\right)\)
\(=\frac{-8}{3^2}.\frac{-15}{4^2}.\frac{-24}{25}...\frac{-2499}{50^2}\)
\(=\frac{8}{3^2}.\frac{15}{4^2}.\frac{24}{5^2}...\frac{2499}{50^2}\) (vì có 48 thừa số âm nên kết quả là dương)
\(=\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}...\frac{49.51}{50.50}\)
\(=\frac{2.3.4...49}{3.4.5...50}.\frac{4.5.6...51}{3.4.5...50}\)
\(=\frac{2}{50}.\frac{51}{3}\)
\(=\frac{1}{25}.17=\frac{17}{25}\)
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Tính D = \(\frac{-1}{3}.\left(1+2+3\right)-\frac{1}{4}.\left(1+2+3\right)-.............-\frac{1}{50}.\left(1+2+3\right)\)
1, Tính \(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)2,Tính \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
\(S=\left(\frac{1}{2}-1\right)\div\left(\frac{1}{3}-1\right)\div\left(\frac{1}{4}-1\right)\div...\div\left(\frac{1}{50}-1\right)\)
vậy S bằng bao nhiêu ?
Ta có S = ( 1/2 - 1) : ( 1/3 - 1) : (1/4 - 1) :... : ( 1/50 - 1)
S = -1/2 : ( -2/3) : ( -3/4) : ... : ( -49/ 50)
S= -1/2 x (-3/2) x ( -4/3) x ... x (-50/49)
S= -1/2 x 1/3 x 50
S= -25/3
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uzumaki naruto trả lời có vẻ kì kì
bấm lộn cái nút
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Xem thêm câu trả lời
\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+.....+\frac{1}{50}\left(1+2+3+...+50\right)\)
Tính tổng
Tìm x biết :
a)\(\frac{2}{3}x-50\%x-\left(-\frac{4}{5}\right):1\frac{3}{5}=-0,12\)\(+1\frac{3}{25}\)
b)\(\left(-1\frac{1}{6}+\frac{2}{3}-\frac{3}{4}\right):x+\left(-1\frac{11}{12}\right).1\frac{21}{23}=-6\frac{1}{3}\)
c)\(50\%x-\frac{1}{3}x-\left(\frac{-2}{3}\right)^2.\left(-1\frac{1}{8}\right)=-119\frac{3}{4}+120\frac{5}{6}\)