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TH
3
N
17 tháng 10 2018
\(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{49}+\frac{1}{50}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{40}+\frac{1}{50}-\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{49}+\frac{1}{50}\)
Vậy .....(tự kết luận)
T
17 tháng 10 2018
CMR: \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}\)
\(VT=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{50}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}-1-\frac{1}{2}-...-\frac{1}{25}\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{50}^{\left(đpcm\right)}\)
UN
2
26 tháng 7 2016
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\left(đpcm\right)\)
S
26 tháng 7 2016
\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{49}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{49}+\frac{1}{50}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\left(đpcm\right)\)
NM
0
VA
3
1 tháng 8 2017
Đặt:
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\)
\(\left\{{}\begin{matrix}\dfrac{1}{2^2}< \dfrac{1}{1.2}\\\dfrac{1}{3^2}< \dfrac{1}{2.3}\\\dfrac{1}{4^2}< \dfrac{1}{3.4}\\\dfrac{1}{50^2}< \dfrac{1}{49.50}\end{matrix}\right.\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(A< 1-\dfrac{1}{50}\)
\(A< 1\rightarrowđpcm\)
PN
1 tháng 8 2017
Ta thấy:
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}< 1-\dfrac{1}{50}\)
\(\Leftrightarrow\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}< \dfrac{49}{50}< 1\)
Vậy \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}< 1\)
LN
1
NT
0
NM
0
TH
1
Vì các p/s bé hơn 1 nên tổng nó bé hơn 1
thế thui
CM: A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{4^2}\)+...+ \(\dfrac{1}{50^2}\) < 1
\(\dfrac{1}{2^2}\) < \(\dfrac{1}{1.2}\) = \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)
\(\dfrac{1}{3^2}\) < \(\dfrac{1}{2.3}\) = \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)
.............................
\(\dfrac{1}{50^2}\) < \(\dfrac{1}{49.50}\) = \(\dfrac{1}{49}\) - \(\dfrac{1}{50}\)
Cộng vế với vế ta có:
A < \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{49}\) - \(\dfrac{1}{50}\)
A < 1 - \(\dfrac{1}{50}\)
A < 1 (đpcm)