Rút gọn tổng sau:
a, A = 1/2+1/2^2+1/2^3+....+1/2^20
b, B = 1/3+1/3^2+1/3^3+....+1/3^21
c, C = 1/1.2.3+1/2.3.4+1/3.4.5+......+1/19.20.21
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U
9 tháng 8 2018
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\right)\)
\(A=1-\frac{1}{2^{20}}\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{21}}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{20}}\)
\(3B-B=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{20}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{21}}\right)\)
\(2B=1-\frac{1}{3^{21}}\)
\(B=\frac{1-\frac{1}{3^{21}}}{2}\)
\(C=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{19\cdot20\cdot21}\)
\(C=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{19\cdot20\cdot21}\right)\)
\(C=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{19\cdot20}-\frac{1}{20\cdot21}\right)\)
\(C=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{20\cdot21}\right)\)
\(C=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{420}\right)\)
\(C=\frac{1}{2}\cdot\frac{209}{420}\)
\(C=\frac{209}{480}\)
TH
13 tháng 2 2018
A = \(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)
3A= \(1+\frac{1}{3}+...+\frac{1}{3^{2006}}+\frac{1}{3^{2007}}\)
3A-A= \(1-\frac{1}{3^{2008}}\)
17 tháng 3 2017
a) A = 1.3 +2.4 + 3.5 +...+ 97.99 + 98.100
A = 1(2 + 1) + 2(3+1) + 3(4 + 1) +...+ 98(99+1)
= (1.2 + 2.3 + 3.4 +...+ 98.99) + (1 + 2 + 3 +...+ 98)
= [ 1.2.3 + 2.3.(4-1) +...+ 98.99.(100-97)] + [ 1.2 + 2.(3-1) + 3.(4-2) +... 98.(99-97)]
= [ 1.2.3 + 2.3.(4-1) - 1.2.3 + 3.4.(5-2) - 2.3.(4-1) +...+ 98.99.(100-97) - 97.98(99-96)] + [ 1.2 + 2.(3-1) - 1.2 + 3.(4-2) - 2.(3-1) +...+ 98.(99-97) - 97(98-96)]
= 98.99.100:3 + 98.99:2 = 323 400 + 4581 = 328251
17 tháng 3 2017
b) B = 1.2.3 + 2.3.4 + 3.4.5 +...+ 48.49.50
4B = 1.2.3.4 + 2.3.4.(5-1) + 3.4.5.(6-2) +...+ 48.49.50.(51-47)
4B-B = 1.2.3.4 + 2.3.4.(5-1) - 1.2.3.4 + 3.4.5.(6-2) - 2.3.4.(5-1) +...+ 48.49.50.(51-47) - 47.48.49.(50-46)
= 48.49.50.51:4 = 1499400
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
20 tháng 7 2023
a/
\(b=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{97.99}\)
\(2b=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+\dfrac{7-5}{5.7}+...+\dfrac{99-97}{97.99}=\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}=\)
\(=1-\dfrac{1}{99}=\dfrac{98}{99}\Rightarrow b=\dfrac{98}{2.99}=\dfrac{49}{99}\)
b/
\(c=\dfrac{3-1}{1.2.3}+\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}=\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+\dfrac{1}{98.99}-\dfrac{1}{99.100}=\)
\(=\dfrac{1}{2}-\dfrac{1}{99.100}\)
c/
\(\dfrac{2}{5}.d=\dfrac{4-2}{2.3.4}+\dfrac{5-3}{3.4.5}+...+\dfrac{100-98}{98.99.100}+\dfrac{101-99}{99.100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}+\dfrac{1}{99.100}-\dfrac{1}{100.101}=\)
\(=\dfrac{1}{2.3}-\dfrac{1}{100.101}\Rightarrow d=\left(\dfrac{1}{2.3}-\dfrac{1}{100.101}\right):\dfrac{2}{5}\)
16 tháng 2 2021
a)
*\(1+2+3+...+\left(n-1\right)+n\)
Số số hạng là:
\(\left(n-1\right):1+1=n-1+1=n\)(số hạng)
Tổng của dãy số là:
\(\left(n+1\right)\cdot\dfrac{n}{2}=\dfrac{n\left(n+1\right)}{2}\)
*\(1+3+5+...+\left(2n-1\right)\)
Số số hạng của dãy số là:
\(\left(2n-1-1\right):2+1=\dfrac{\left(2n-2\right)}{2}+1=n-1+1=n\)(số hạng)
Tổng của dãy số là:
\(\left(2n-1+1\right)\cdot\dfrac{n}{2}=\dfrac{2n^2}{2}=2n\)