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\(B=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{50}\cdot\dfrac{50\cdot51}{2}\)
\(=1+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{51}{2}\)
\(=\dfrac{50\cdot\dfrac{\left(51+2\right)}{2}}{2}=50\cdot\dfrac{53}{4}=662.5\)
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 )
\(=\left(1\frac{1}{2}+\frac{1}{2}\right)+\left(2\frac{2}{3}+\frac{1}{3}\right)+...+\left(50\frac{50}{51}+\frac{1}{51}\right)\)
\(=2+3+...+51\)
\(=\frac{\left(2+51\right)50}{2}\)
\(=1325\)
Đặt \(S=\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+.....+99}+\frac{1}{50}\)
Đặt E = \(\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+....+99}\)
\(E=\frac{1}{2.3:2}+\frac{1}{3.4:2}+....+\frac{1}{99.100:2}\)
\(\frac{1}{2}E=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
E = 49/100 : 1/2 = 49/50
Vậy \(S=\frac{49}{50}+\frac{1}{50}=\frac{50}{50}=1\)
\(P=1+\frac{1}{1+2}+\frac{1}{1+2+3}+........+\frac{1}{1+2+3+.......+50}=1+\frac{1}{\frac{2.3}{2}}+\frac{1}{\frac{3.4}{2}}+......+\frac{1}{\frac{50.51}{2}}=1+\frac{2}{2.3}+\frac{2}{3.4}+......+\frac{2}{50.51}=1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+......+\frac{1}{50.51}\right)\) \(Taco:\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\)
\(\Rightarrow P=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.......+\frac{1}{50}-\frac{1}{51}\right)=1+2\left(\frac{1}{2}-\frac{1}{51}\right)=1+1-\frac{2}{51}=2-\frac{2}{51}=\frac{100}{51}\)
\(1\dfrac{1}{2}+2\dfrac{2}{3}+3\dfrac{3}{4}+...+50\dfrac{50}{51}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{51}\)
\(=\left(1\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3\dfrac{3}{4}+\dfrac{1}{4}\right)+...+\left(50\dfrac{50}{51}+\dfrac{1}{51}\right)\)
\(=2+3+4+...+51\)
\(=\dfrac{50\left(51+2\right)}{2}\)
=1325
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
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 )
\(=1275+50\)
\(=1325\)