\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.........+\frac{1}{1+2+3+...+50}\)
Những câu hỏi liên quan
Tính \(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}+4\frac{4}{5}+...+50\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\)
Từ dãy trên ta có:
(\(\frac{3}{2}\)+\(\frac{1}{2}\))+(\(\frac{8}{3}\)+\(\frac{2}{3}\))+......+(\(\frac{2600}{51}\)+\(\frac{1}{51}\)) < vì không có cách nhập hỗn số nên mình đổi ra phân số >
= 2 + 3 + 4 + 5 + 6 + ..........................+ 51
Từ 2 -> 51 có :( 51 - 2 ) : 1 + 1 = 50 số
Chia ra : 50 : 2 = 25 cặp
ta có( 51 + 2 ) x 25 =1325
Vậy tổng trên có kết quả bằng 1325 (tớ chỉ nghĩ thế thôi chứ sai đừng trách nhá.Đùa thôi,đúng đấy )
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Tính \(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}+4\frac{4}{5}+...+50\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\)
\(=\left(1\frac{1}{2}+\frac{1}{2}\right)+\left(2\frac{2}{3}+\frac{1}{3}\right)+...+\left(50\frac{50}{51}+\frac{1}{51}\right)\)
\(=2+3+...+51\)
\(=\frac{\left(2+51\right)50}{2}\)
\(=1325\)
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Câu 8:=
1\(\frac{1}{2}\)+2\(\frac{2}{3}\)+3\(\frac{3}{4}\)+4\(\frac{4}{5}\)+.......+50\(\frac{50}{51}\)+\(\frac{1}{2}\)+\(\frac{1}{3}\)+\(\frac{1}{4}\)+\(\frac{1}{5}\)+....+\(\frac{1}{51}\)
=(1\(\frac{1}{2}\)+\(\frac{1}{2}\))+(2\(\frac{2}{3}\)+\(\frac{1}{3}\))+(3\(\frac{3}{4}\)+\(\frac{1}{4}\))+.......+(50\(\frac{50}{51}\)+\(\frac{1}{51}\))
=2+3+4+.....+51
=1325
Vậy:1\(\frac{1}{2}\)+2\(\frac{2}{3}\)+3\(\frac{3}{4}\)+4\(\frac{4}{5}\)+.......+50\(\frac{50}{51}\)+\(\frac{1}{2}\)+\(\frac{1}{3}\)+\(\frac{1}{4}\)+\(\frac{1}{5}\)+....+\(\frac{1}{51}\)=1325
Học Tốt!
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\(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}+4\frac{4}{5}+...+50\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{51}\)
\(=1+\frac{1}{2}+2+\frac{2}{3}+3+\frac{3}{4}+...+50+\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\)
\(=\left(1+2+3+...+50\right)+\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{2}{3}+\frac{1}{3}\right)+...+\left(\frac{50}{51}+\frac{1}{51}\right)\)
\(=\frac{50.51}{2}+1+1+1+...+1\) ( có 50 số 1 )
\(=1275+50\)
\(=1325\)
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cứ nhóm vào ta được
2+3+......+50+51
suy ra biểu thức trên bằng 1325
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Tính \(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}+4\frac{4}{5}+...+50\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\)
\(1\dfrac{1}{2}+2\dfrac{2}{3}+3\dfrac{3}{4}+...+50\dfrac{50}{51}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{51}\)
\(=\left(1\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3\dfrac{3}{4}+\dfrac{1}{4}\right)+...+\left(50\dfrac{50}{51}+\dfrac{1}{51}\right)\)
\(=2+3+4+...+51\)
\(=\dfrac{50\left(51+2\right)}{2}\)
=1325
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\(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}+...+50\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...\frac{1}{51}\)
Ta có :
\(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}+...+50\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\)
= \(\left(1\frac{1}{2}+\frac{1}{2}\right)+\left(2\frac{2}{3}+\frac{1}{3}\right)+\left(3\frac{3}{4}+\frac{1}{4}\right)+...+\left(49\frac{49}{50}+\frac{1}{50}\right)+\left(50\frac{50}{51}+\frac{1}{51}\right)\)
= \(2+3+4+5+...+49+50+51\)
= \(\left(\frac{51-2}{1}+1\right).\frac{51+2}{2}\)
= \(50.26,5\)
= 1325
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frac{1}{2^2}+frac{1}{4^2}+frac{1}{6^2}+..+frac{1}{100^2}frac{1}{2^2}left(1+frac{1}{2^2}+frac{1}{3^2}+...+frac{1}{50^2}right)Có frac{1}{2^2} frac{1}{1.2}1-frac{1}{2} frac{1}{3^2} frac{1}{2.3}frac{1}{2}-frac{1}{3}....v........v............ frac{1}{50^2} frac{1}{49.50}frac{1}{49}-frac{1}{50}Cộng lại 1+frac{1}{2^2}+frac{1}{3^2}+...+frac{1}{50^2} 1+1-frac{1}{2}+frac{1}{2}-frac{1}{3}+...+frac{1}{49}-frac{1}{50}2-frac{1}{50}Rightarrow VT frac{1}{2^2}left(2-frac{1}{50}right)frac{1}{2...
Đọc tiếp
\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+..+\frac{1}{100^2}=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
Có \(\frac{1}{2^2}< \frac{1}{1.2}=1-\frac{1}{2}\) \(\frac{1}{3^2}< \frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)....v........v............ \(\frac{1}{50^2}< \frac{1}{49.50}=\frac{1}{49}-\frac{1}{50}\)
Cộng lại \(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=2-\frac{1}{50}\)
\(\Rightarrow VT< \frac{1}{2^2}\left(2-\frac{1}{50}\right)=\frac{1}{2}-\frac{1}{2^2.50}< \frac{1}{2}\left(Đpcm\right)\)
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ủa toán lớp mấy chứ ko phải lớp 1
uk ko phải toán lớp 1
Tính: \(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}+...+50\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{51}\)= ________?
Tính tổng :\(S=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+....+\frac{1}{50}.\left(1+2+3+4+....+50\right)\)
Cho day phan so:
\(\frac{1}{1};\frac{1}{2};\frac{2}{1};\frac{1}{3};\frac{2}{2};\frac{3}{1};\frac{1}{4};\frac{2}{3};\frac{3}{2};\frac{4}{1}\)
Tim ra phan so 50 cua day so tren
e, \(\frac{49}{1}+\frac{48}{2}+\frac{47}{3}+.......+\frac{2}{48}+\frac{1}{49}=50.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{50}\right)\)
\(\frac{49}{1}+\frac{48}{2}+\frac{47}{3}+...+\frac{2}{48}+\frac{1}{49}\)
\(=1+1+...+1+\frac{48}{2}+\frac{47}{3}+...+\frac{2}{48}+\frac{1}{49}\)(có 49 số 1)
\(=\left(1+\frac{48}{2}\right)+\left(1+\frac{47}{3}\right)+...+\left(1+\frac{2}{48}\right)+\left(1+\frac{1}{49}\right)+1\)
\(=\frac{50}{2}+\frac{50}{3}+...+\frac{50}{48}+\frac{50}{49}+\frac{50}{50}\)
\(=50\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{50}\right)\)
Chúc bạn học tốt.
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