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2 tháng 3 2017
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 )
BT
1
TT
3 tháng 10 2023
E = -1/3 +1/(3^2) - 1/(3^3) + .... - 1/(3^51)
E.1/3 = -1/(3^2) + 1/(3^3)-1/(3^4) +.... - 1/(3^52)
E + E.1/3 = [-1/3+1/(3^2) - 1/(3^3) +.... -1/(3^51)]+[-1/(3^2) +1/(3^3) -1/(3^4) +.... - 1/(3^52)]
E.4/3 = -1/3-1/(3^52)
E.4/3 = (-3^51 - 1)/(3^52)
E = (-3^51 - 1)/(3^52) . 3/4
E = (-3^51-1)/(4.3^51)
28 tháng 2 2017
\(=\left(1\frac{1}{2}+\frac{1}{2}\right)+\left(2\frac{2}{3}+\frac{1}{3}\right)+...+\left(50\frac{50}{51}+\frac{1}{51}\right)\)
\(=2+3+...+51\)
\(=\frac{\left(2+51\right)50}{2}\)
\(=1325\)
M
3 tháng 3 2016
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
BH
27 tháng 2 2017
\(1\dfrac{1}{2}+2\dfrac{2}{3}+3\dfrac{3}{4}+...+50\dfrac{50}{51}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{51}\)
\(=\left(1\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3\dfrac{3}{4}+\dfrac{1}{4}\right)+...+\left(50\dfrac{50}{51}+\dfrac{1}{51}\right)\)
\(=2+3+4+...+51\)
\(=\dfrac{50\left(51+2\right)}{2}\)
=1325
14 tháng 8 2023
\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{1\cdot2\cdot3}+\dfrac{2}{2\cdot3\cdot4}+...+\dfrac{2}{50\cdot51\cdot52}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{50\cdot51}-\dfrac{1}{51\cdot52}\right)\)
\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{51\cdot52}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{1325}{2652}=\dfrac{1325}{5304}\)
19 tháng 2 2017
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!
QD
20 tháng 2 2017
\(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}+4\frac{4}{5}+...+50\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{51}\)
\(=1+\frac{1}{2}+2+\frac{2}{3}+3+\frac{3}{4}+...+50+\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\)
\(=\left(1+2+3+...+50\right)+\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{2}{3}+\frac{1}{3}\right)+...+\left(\frac{50}{51}+\frac{1}{51}\right)\)
\(=\frac{50.51}{2}+1+1+1+...+1\) ( có 50 số 1 )
\(=1275+50\)
\(=1325\)
M
1
Đặt \(A=\dfrac{1}{3}+\dfrac{1}{3^3}+\dfrac{1}{3^5}+...+\dfrac{1}{3^{47}}+\dfrac{1}{3^{49}}+\dfrac{1}{3^{51}}\)
Và \(B=\dfrac{1}{3^2}+\dfrac{1}{3^4}+\dfrac{1}{3^6}+...+\dfrac{1}{3^{46}}+\dfrac{1}{3^{48}}+\dfrac{1}{3^{50}}\)
Ta có:
\(9A=3+\dfrac{1}{3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{45}}+\dfrac{1}{3^{47}}+\dfrac{1}{3^{49}}\)
\(9A-A=\left(3+\dfrac{1}{3}+...+\dfrac{1}{3^{47}}+\dfrac{1}{3^{49}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{49}}+\dfrac{1}{3^{51}}\right)\)
\(8A=3-\dfrac{1}{3^{51}}\)
\(A=\dfrac{3-\dfrac{1}{3^{51}}}{8}\)
\(9B=1+\dfrac{1}{3^2}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{44}}+\dfrac{1}{3^{46}}+\dfrac{1}{3^{48}}\)
\(9B-B=\left(1+\dfrac{1}{3^2}+...+\dfrac{1}{3^{46}}+\dfrac{1}{3^{48}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{48}}+\dfrac{1}{3^{50}}\right)\)
\(8B=1-\dfrac{1}{3^{50}}\)
\(B=\dfrac{1-\dfrac{1}{3^{50}}}{8}\)
Suy ra
\(-\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{50}}-\dfrac{1}{3^{51}}=B-A=\dfrac{1-\dfrac{1}{3^{50}}}{8}-\dfrac{3-\dfrac{1}{3^{51}}}{8}\)
\(=\dfrac{\left(1-\dfrac{1}{3^{50}}\right)-\left(3-\dfrac{1}{3^{51}}\right)}{8}=\dfrac{-2-\dfrac{1}{3^{50}}+\dfrac{1}{3^{51}}}{8}=\dfrac{-2+\dfrac{-3^{51}+3^{50}}{3^{101}}}{8}\)
\(=\dfrac{-2+\dfrac{3^{50}\left(-3+1\right)}{3^{101}}}{8}=\dfrac{-2-\dfrac{2}{3^{51}}}{8}=-\dfrac{2\left(1+\dfrac{1}{3^{51}}\right)}{8}=-\dfrac{1+\dfrac{1}{3^{51}}}{4}\)