Tính tổng :
a) \(C=1\cdot2+2\cdot3+...+100\cdot101\)
b) \(A=1\cdot2+2\cdot3+...+n\left(n+1\right)\)
Những câu hỏi liên quan
Tính tổng :a) Afrac{5}{2cdot1}+frac{4}{1cdot11}+frac{3}{11cdot14}+frac{1}{14cdot15}+frac{13}{15cdot28}b) Bfrac{-1}{20}+frac{-1}{30}+frac{-1}{42}+frac{-1}{56}+frac{-1}{72}+frac{-1}{90}c) Cfrac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{nleft(n+1right)left(n+2right)}d) Dfrac{1}{1cdot2cdot3cdot4}+frac{1}{2cdot3cdot4cdot5}+...+frac{1}{nleft(n+1right)left(n+2right)left(n+3right)}e) Eleft(frac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{37cdot38cdot39}right)cdot1482cdot185cdot8
Đọc tiếp
Tính tổng :
a) \(A=\frac{5}{2\cdot1}+\frac{4}{1\cdot11}+\frac{3}{11\cdot14}+\frac{1}{14\cdot15}+\frac{13}{15\cdot28}\)
b) \(B=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
c) \(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
d) \(D=\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
e) \(E=\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\right)\cdot1482\cdot185\cdot8\)
Tính \(\frac{B}{A}\)biết
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+....+\frac{1}{n\left(n+1\right)}+...+\frac{1}{2008\cdot2009}\)
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}+...+\frac{1}{2008\cdot2009\cdot2010}\)
Ta có
\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\) và \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n+2}\) nên
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{n\left(n+1\right)}+...+\frac{1}{2008\cdot2009}=1-\frac{1}{2009}=\frac{2008}{2009}\)
\(2B=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}+...+\frac{2}{2008\cdot2009\cdot2010}\)
\(=\frac{1}{1\cdot2}-\frac{1}{2009\cdot2010}=\frac{201944}{2009\cdot2010}\)
\(\Rightarrow B=\frac{1}{2}\cdot\frac{201944}{2009\cdot2010}=\frac{1009522}{2009\cdot2010}\)
Do đó \(\frac{B}{A}=\frac{1009522}{2009\cdot2010}:\frac{2008}{2009}=\frac{1009522\cdot2009}{2008\cdot2009\cdot2010}=\frac{5047611}{2018040}\)
Tính tổng :
a) Afrac{5}{2cdot1}+frac{4}{1cdot11}+frac{3}{11cdot14}+frac{1}{14cdot15}+frac{13}{15cdot28}
b) Bfrac{-1}{20}+frac{-1}{30}+frac{-1}{42}+frac{-1}{56}+frac{-1}{72}+frac{-1}{90}
c) Cfrac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{nleft(n+1right)left(n+2right)}
d) Dfrac{1}{1cdot2cdot3cdot4}+frac{1}{2cdot3cdot4cdot5}+...+frac{1}{nleft(n+1right)left(n+2right)left(n+3right)}
e) Eleft(frac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{37cdot38cdot39}right)cdot1482cdot185cdot8
Đọc tiếp
Tính tổng :
a) \(A=\frac{5}{2\cdot1}+\frac{4}{1\cdot11}+\frac{3}{11\cdot14}+\frac{1}{14\cdot15}+\frac{13}{15\cdot28}\)
b) \(B=\frac{-1}{20}+\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
c) \(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
d) \(D=\frac{1}{1\cdot2\cdot3\cdot4}+\frac{1}{2\cdot3\cdot4\cdot5}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}\)
e) \(E=\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\right)\cdot1482\cdot185\cdot8\)
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{A}{7}=\frac{7-2}{2.7}+\frac{11-7}{7.11}+\frac{14-11}{11.4}+\frac{15-14}{14.15}+\frac{28-15}{15.28}\)
\(\frac{A}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}=\frac{1}{2}-\frac{1}{28}=\frac{13}{28}\)
\(A=7.\frac{13}{28}\)
\(A=\frac{13}{4}\)
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TÍnh Tổng sau : \(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot3}+...+\frac{1}{a+\left(a+1\right)+\left(a+2\right)}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{n.\left(n+1\right).\left(n+2\right)}\)
\(=\frac{n.\left(n+3\right)}{4.\left(n+1\right).\left(n+2\right)}\)
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Bạn ghi rõ cách làm cho mình đc ko minh ko hiểu
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ok ok , đề bài sai : Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{a.\left(a+1\right).\left(a+2\right)}\)
\(A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{a.\left(a+1\right)}-\frac{1}{\left(a+1\right).\left(a+2\right)}\)
\(A=\frac{1}{1.2}-\frac{1}{\left(a+1\right).\left(a+2\right)}=\frac{1}{2}-\frac{1}{\left(a+1\right).\left(a+2\right)}\)
Vậy \(A=\frac{1}{2}-\frac{1}{\left(a+1\right).\left(a+2\right)}\)
Cơ bản thì là vậy , nhưng cái phần tách ra mình ko nhớ , mình nhớ ra sẽ thông báo sao nhé
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Xem thêm câu trả lời
TÍNH TỔNG:
\(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+.....+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
Bài 1:
a) \(\frac{1}{1}\cdot2+\frac{1}{2}\cdot3+\frac{1}{3}\cdot4+...+\frac{1}{n}\cdot\left(n+1\right)\)
b) \(\frac{1}{1}\cdot2\cdot3+\frac{1}{2}\cdot3\cdot4+\frac{1}{3}\cdot4\cdot5+...+\frac{1}{a}\cdot\left(a+1\right)\cdot\left(a+2\right)\)
tính tổng sau:
\(A=1\cdot2\cdot3\cdot.....\cdot2021-1\cdot2\cdot3\cdot.....\cdot2021-1\cdot2\cdot3\cdot.....\cdot2019\cdot2020^2\)
\(A=1.2.3.4...2019.\left(2020.2021-2020^2\right)=1.2.3.4...2019.2020\)
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chứng minh rằng :
a) \(a^2+b^2+c^2-ab-bc-ca=0\) thì \(a=b=c\)
b)\(a^2+b^2+c^2\ge ab+bc+ca\)Với mọi a,b,c
c)\(\left(3^{n+1}-2\cdot2^n\right)\left(3\cdot3^n+2^{n+1}\right)\cdot3^{2n+2}+\left(8\cdot2^{n-2}\cdot3^{n+1}\right)^2\)là một số chính phương với mọi số tự nhiên n
a^2 + b^2 + c^2= ab + bc + ca
2 ( a^2 + b^2 + c^2 ) = 2 ( ab + bc + ca)
2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca
a^2 + a^2 + b^2 + b^2 + c^2+ c^2 – 2ab – 2bc – 2ca = 0
a^2 + b^2 – 2ab + b^2 + c^2 – 2bc + c² + a² – 2ca = 0
(a^2 + b^2 – 2ab) + (b^2 + c^2 – 2bc) + (c^2 + a^2 – 2ca) = 0
(a – b)^2 + (b – c)^2 + (c – a)^2 = 0
Vì (a-b)^2 lớn hơn hoặc bằng 0 với mọi a và b
(b-c)^2 lớn hơn hoặc bằng 0 với mọi c và b
(c-a)^2 lớn hơn hoặc bằng 0 với mọi a và c
=> (a-b)^2 =0 ; (b-c)^2=0 ; (c-a)^2=0
=> a=b ; b=c ; c=a
=>a=b=c
\(101\cdot\left(\dfrac{1}{1\cdot2}+\dfrac{5}{2\cdot3}+\dfrac{11}{3\cdot4}+.....+\dfrac{10099}{100\cdot101}\right)\)
Thanks các bạn nhìu