Tính
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{50}\)
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 )
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 :\(S=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+....+\frac{1}{50}.\left(1+2+3+4+....+50\right)\)
Tính A = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}}{\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+\frac{49}{1}}\)
A = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}}{\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+\frac{49}{1}}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}}{\left(\frac{1}{49}+1\right)+\left(\frac{2}{48}+1\right)+\left(\frac{3}{47}+1\right)+...+\left(\frac{48}{2}+1\right)+\frac{50}{50}}\)
A = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}}{\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+...+\frac{50}{2}+\frac{50}{50}}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}}{\left(\frac{1}{49}+\frac{1}{48}+\frac{50}{47}+...+\frac{1}{2}+\frac{1}{50}\right).50}=\frac{1}{50}\)
\(A=\frac{T}{M}\)
\(M=\frac{1}{49}+1+\frac{2}{48}+1+\frac{3}{47}+1+.........+\frac{48}{2}+1+1\)
\(=\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+.........+\frac{50}{2}+1\)
\(=50.\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+......+\frac{1}{2}+\frac{1}{50}\right)=50.T\)
\(A=\frac{T}{50T}=\frac{1}{50}\)
Tính \(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}+4\frac{4}{5}+...+50\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\)
\(=\left(1\frac{1}{2}+\frac{1}{2}\right)+\left(2\frac{2}{3}+\frac{1}{3}\right)+...+\left(50\frac{50}{51}+\frac{1}{51}\right)\)
\(=2+3+...+51\)
\(=\frac{\left(2+51\right)50}{2}\)
\(=1325\)
tính
\(-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-....+\frac{1}{3^{50}}-\frac{1}{3^{51}}\)
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!
\(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}+4\frac{4}{5}+...+50\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{51}\)
\(=1+\frac{1}{2}+2+\frac{2}{3}+3+\frac{3}{4}+...+50+\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\)
\(=\left(1+2+3+...+50\right)+\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{2}{3}+\frac{1}{3}\right)+...+\left(\frac{50}{51}+\frac{1}{51}\right)\)
\(=\frac{50.51}{2}+1+1+1+...+1\) ( có 50 số 1 )
\(=1275+50\)
\(=1325\)
cứ nhóm vào ta được
2+3+......+50+51
suy ra biểu thức trên bằng 1325
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
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.....+\frac{1}{1+2+3+4+....+50}\)
Tính tổng trên
1. Tính hợp lý
\(\frac{\frac{1}{9}-\frac{5}{6}-4}{\frac{7}{12}-\frac{1}{36}-10}\)
2. CMr:
\(\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
2) 1\26+1\27+1\28+........+1\50=1+1\2+1\3+......+1\50 -( 1+1\2+1\3+.....+1\25)=1+1\2+1\3+....+1\50-2.(1\2+1\4+1\6+....+1\50)=1-1\2+1\3-1\4+.....+1\49-1\50=vế phải(đpcm)