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ủa toán lớp mấy chứ ko phải lớp 1

uk ko phải toán lớp 1

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 )

Bài 1:

\(\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)

= \(\left[\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\right]\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)

= \(\left[\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\right]\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)

=\(\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\right]\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)

=\(\frac{1}{26}+\frac{1}{27}+....+\frac{1}{26}\):\(\left(\frac{1}{25}+\frac{1}{26}+....+\frac{1}{50}\right)\)

......????

\(=\left(1\frac{1}{2}+\frac{1}{2}\right)+\left(2\frac{2}{3}+\frac{1}{3}\right)+...+\left(50\frac{50}{51}+\frac{1}{51}\right)\)

\(=2+3+...+51\)

\(=\frac{\left(2+51\right)50}{2}\)

\(=1325\)

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

\(1\dfrac{1}{2}+2\dfrac{2}{3}+3\dfrac{3}{4}+...+50\dfrac{50}{51}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{51}\)

\(=\left(1\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3\dfrac{3}{4}+\dfrac{1}{4}\right)+...+\left(50\dfrac{50}{51}+\dfrac{1}{51}\right)\)

\(=2+3+4+...+51\)

\(=\dfrac{50\left(51+2\right)}{2}\)

=1325

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

\(y=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+....+\frac{1}{1+2+3+...+49+50}=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{1275}\)\(=2\cdot\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2550}\right)=2\cdot\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{50.51}\right)\)\(=2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)=2\cdot\left(\frac{1}{2}-\frac{1}{51}\right)=2\cdot\frac{49}{102}=\frac{49}{51}\)

kq cuối nk =1326 (vừa nhìn nhầm )

=2550 nha (hình như thế) 

nhầm =2601