cho biểu thức:A=\(\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+...\dfrac{ }{ }\)\(\dfrac{1}{40}\)
CM:A>\(\dfrac{1}{2}\)
người nào giải đầu tiên mà nhanh mik sẽ tick
Nếu 1 + \(\dfrac{1}{1-\dfrac{1}{2}}\) + \(\dfrac{1}{1-\dfrac{2}{3}}\) + \(\dfrac{1}{1-\dfrac{3}{4}}\) + .... + \(\dfrac{1}{1+\dfrac{n}{n+1}}\) = 276, thế n sẽ là gì ?
A) 21 B) 22 C) 23 D)24 E) 25
cả dãy đang trừ mà sao cái cuối là cộng vậy bạn, dãy ko có quy tắc à :v
Hmm đề sai :v
Sửa lại:
`1+1/(1-1/2)+1/(1-/3)+1/(1-3/4)+...........+1/(1+n/(n+1))=276`
`=>1+1/(1/2)+1/(1/3)+1/(1/4)+......+1/(1/n)=276`
`=>1+2+3+4+.......+n=276`
Từ `1->n` có n số
`=>1+2+3+4+.....+n=(n(n+1))/2`
`=>(n(n+1))/2=276`
`=>n(n+1)=552`
`=>n(n+1)=23.24`
`=>n=23`
Vậy `n=23`
a , cho A = \(\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}\) . Chứng minh A < \(\dfrac{7}{4}\)
b ,cho B = 21 + 22 + 23 + ... + 260 . Chứng minh B \(⋮\) 21
b.ta chia B thành 10 nhóm mỗi nhóm có 6 hạng tử \(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+....+\left(2^{55}+2^{56}+2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(B\text{=}2\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{55}\left(1+2+2^2+2^3+2^4+2^5\right)\)
\(B\text{=}2.63+...+2^{56}.63\)
\(\Rightarrow B⋮63\)
\(\Rightarrow B⋮21\)
So sánh phân số \(\dfrac{2012}{2013}\) với A, biết:
A = \(\dfrac{1}{11}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{13}\) + \(\dfrac{1}{14}\) + \(\dfrac{1}{15}\) + ... + \(\dfrac{1}{38}\) + \(\dfrac{1}{39}\) + \(\dfrac{1}{40}\)
Ai xong đầu tiên mình tick nhé.
CTR\(\dfrac{11}{15}< \dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+...+\dfrac{1}{59}+\dfrac{1}{60}< \dfrac{3}{2}\)
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Chứng tỏ rằng :
\(\dfrac{7}{12}\)<\(\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+............+\dfrac{1}{40}\)<\(\dfrac{5}{6}\)
Tìm x biết:
\(\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{11}{40},\left(x\inℕ^∗\right)\)
Giải chi tiết giúp mik nha.
\(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\) + \(\dfrac{1}{36}\) +...+ \(\dfrac{2}{x\left(x+1\right)}\) = \(\dfrac{11}{40}\) (\(x\in\) N*)
\(\dfrac{1}{2}\).(\(\dfrac{1}{15}\)+\(\dfrac{1}{21}\)+\(\dfrac{1}{28}\)+\(\dfrac{1}{36}\)+.....+ \(\dfrac{2}{x\left(x+1\right)}\)) = \(\dfrac{11}{40}\) \(\times\) \(\dfrac{1}{2}\)
\(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)+...+ \(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\)+...+ \(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+...+ \(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{5}\) - \(\dfrac{1}{x+1}\) = \(\dfrac{11}{80}\)
\(\dfrac{1}{x+1}\) = \(\dfrac{1}{5}\) - \(\dfrac{11}{80}\)
\(\dfrac{1}{x+1}\) = \(\dfrac{1}{16}\)
\(x\) + 1 = 16
\(x\) = 16 - 1
\(x\) = 15
\(\dfrac{29-x}{21}\)+\(\dfrac{27-x}{23}\)+\(\dfrac{25-x}{25}\)+\(\dfrac{23-x}{27}\)+\(\dfrac{21-x}{29}\)=\(\dfrac{(29-x+1}{21}\)+\(\dfrac{(27-x+1)}{23}\)+\(\dfrac{(25-x+1)}{25}\)+\(\dfrac{(23-x+1)}{21}\)=-5 +5
GIẢI nốt hộ mình với ạ
Cho A = \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+\dfrac{1}{44}+......+\dfrac{1}{80}\)
Chứng tỏ \(\dfrac{7}{12}< A< \dfrac{5}{6}\)
Mik sẽ tick cho ai giải nhanh , đúng và đầy đủ nhất nha !
SOS ! Help me
(\(\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{30}\) ).(\(\dfrac{21}{22}+\dfrac{22}{23}+......+\dfrac{102}{103}\))
\(\left(\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{30}\right).\left(\dfrac{21}{22}+\dfrac{22}{23}+...+\dfrac{102}{103}\right)\)
\(=0.\left(\dfrac{21}{22}+\dfrac{22}{23}+...+\dfrac{102}{103}\right)\)
\(=0\)
vì \(\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{1}{30}\)=0 nên
(15−16−13015−16−130 ).(2122+2223+......+102103)=0