1/7+(1/7)^2+(1/7)^3+(1/7)^4+......+(1/7)^99+(1/7)^100
Những câu hỏi liên quan
Tính:
a)1*4*7+4*7*10+7*10*13+....+100*103*106
b)1*4+4*7+7*10+.....+100*103
c)1*1*1+4*4*4+7*7*7+....+99*99*99
d)1*3*3*3+3*5*5*5+5*7*7*7+.....+49*51*51*51
e)1*99+2*98+3*97+......+50*50
f)1*99+3*97+5*95+....+49*51
Giúp mình nhé!
\(\dfrac{1}{7^2}+\dfrac{1}{7^3}+\dfrac{1}{7^4}+...+\dfrac{1}{7^{99}}+\dfrac{1}{7^{100}}\)
Đặt \(A=\dfrac{1}{7^2}+\dfrac{1}{7^3}+...+\dfrac{1}{7^{100}}\)
\(7A=\dfrac{1}{7}+\dfrac{1}{7^2}+...+\dfrac{1}{7^{99}}\)
\(\Rightarrow7A-A=\dfrac{1}{7}-\dfrac{1}{7^{100}}\)
\(\Rightarrow6A=\dfrac{1}{7}-\dfrac{1}{7^{100}}\)
\(\Rightarrow A=\dfrac{1}{6}\left(\dfrac{1}{7}-\dfrac{1}{7^{100}}\right)\)
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2 tháng 4 2023 lúc 19:23
so sánh
P=\(\dfrac{1+7^2+7^3+...+7^{100}}{1+7^2+7^3+...+7^{99}}\)
Q=\(\dfrac{1+9^2+9^3+...+9^{100}}{1+9^2+9^3+...+9^{99}}\)
Tính:
A=1+7+7^2 +7^3+..+7^2007
B=1+4+4^2+4^3+...+4^100
C=1+3^2+3^4+3^6+3^8+...+3^100
D=7+7^3+7^5+7^7+7^9+...+7^99
E=2+2^3+2^5+2^7+2^9+...+2^9009
\(A=1+7+7^2+7^3+...+7^{2007}\)
\(7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(7A-A=\left(7+7^2+7^3+7^4+...+7^{2008}\right)-\left(1+7+7^2+7^3+...+7^{2007}\right)\)
\(6A=7^{2008}-1\)
\(A=\frac{7^{2008}-1}{6}\)
Tương tự, \(B=\frac{4^{101}-1}{3},C=\frac{3^{101}-1}{2}\).
\(D=7+7^3+7^5+7^7+...+7^{99}\)
\(7^2.D=7^3+7^5+7^7+7^9+...+7^{101}\)
\(\left(7^2-1\right)D=\left(7^3+7^5+7^7+7^9+...+7^{101}\right)-\left(7+7^3+7^5+7^7+...+7^{99}\right)\)
\(48D=7^{101}-7\)
\(D=\frac{7^{101}-7}{48}\)
Tương tự, \(E=\frac{2^{9011}-2}{3}\)
cho t =1/7^2 +2/7^3+3/7^4+....+99/7^100 chứng minh t < 1/36
Lời giải:
$T = \frac{1}{7^2}+\frac{2}{7^3}+\frac{3}{7^4}+....+\frac{99}{7^{100}}$
$7T = \frac{1}{7}+\frac{2}{7^2}+\frac{3}{7^3}+....+\frac{99}{7^{99}}$
$\Rightarrow 6T=7T-T = \frac{1}{7}+\frac{1}{7^2}+\frac{1}{7^3}+...+\frac{1}{7^{99}}-\frac{99}{7^{100}}$
$42T = 1+\frac{1}{7}+\frac{1}{7^2}+...+\frac{1}{7^{98}}-\frac{99}{7^{99}}$
$\Rightarrow 42T-6T = 1-\frac{100}{7^{99}}+\frac{99}{7^{100}}$
$\Rightarrow 36T = 1-\frac{601}{7^{100}}< 1$
$\Rightarrow T< \frac{1}{36}$
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So sánh A và B biết : A= 1+7+7^2 +......+7^100 / 1 + 7 + 7^2 +..... +7^99 ; B = 1 + 9 + 9^2 + 9^3 +......+9^100 / 1+9+9^2+9^99
A= 1/1×2+1/2×3+...1/98×99+1/99×100
B=4/3×7+4/7×11+4/11×15+...4/107×111
C=7/10×11+7/11×12+7/12×13+...7/69×70
Các bạn làm ơn giúp mình với
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
\(A=\frac{99}{100}\)
\(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{107}-\frac{1}{111}\)
\(B=\frac{1}{3}-\frac{1}{111}\)
\(B=\frac{12}{37}\)
\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(C=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(C=7\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(C=7.\frac{3}{35}\)
\(C=\frac{3}{5}\)
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Ta có:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)
\(B=4.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\right)\)
\(B=4.\left(\frac{1}{3}-\frac{1}{111}\right)=4.\frac{12}{37}=\frac{48}{37}\)
\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(C=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(C=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)
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\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}=\frac{99}{100}\)
\(B=\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{107.111}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{107}-\frac{1}{111}\)
\(B=\frac{1}{3}-\frac{1}{111}=\frac{12}{37}\)
\(C=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(C=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(C=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}\)
\(\Rightarrow C=\frac{3}{5}\)
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1)11-12+13-14+15-16+17-18+19-20+21-22+.........+99-100
2)2-4+6-8+......+1998-2000
3)-1+3-5+7-....+97-99
4)1+2-3-4+.........+97+98-99-100
5)1-2+3-4+.............+99-100
6)1+3-5-7+......+97-98-99+100
7)2100-299-298-..........22-2-1
8)1-4+7-10+........+307-310+313
tính các tổng sau
1) A = 1+7+7^2+7^3+....+7^2007
2) B= 1+4 +4^2+4^3+....+4^100
3) C= 1+3^2 +3^4 +3^6+3^8+....+3^100
4) D= 7+7^3 + 7^5+7^7+7^9+....+7^99
5)E= 2+2^3+2^5+2^7+2^9+....+2^2009
6) B = 1+2^2+2^4+2^6+2^8+....+2^200
7) C= 5+5^3+5^5+5^9+....+5^101
8) D = 13+13^3+13^5+...+13^99
Mình làm mẫu 1 bài rùi bạn tự giải những bài còn lại nha
1, 7A = 7+7^2+7^3+....+7^2008
6A = 7A - A = (7+7^2+7^3+....+7^2008)-(1+7+7^2+....+7^2007) = 7^2008-1
=> A = (7^2008-1)/6
Tk mk nha
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\(A=1+7+7^2+7^3+...+7^{2007}\)
\(\Rightarrow7A=7+7^2+7^3+7^4+...+7^{2008}\)
\(\Rightarrow7A-A=\left(7+7^2+7^3+...+7^{2008}\right)-\left(1+7+7^2+...+7^{2007}\right)\)
\(\Rightarrow6A=7^{2008}-1\)
\(\Rightarrow A=\frac{7^{2008}-1}{6}\)
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4b=4+4^2+4^3+...+4^101
4b-b=(4+4^2+...+4^101)-(1+4+4^2+...+4^100)
3b=4^101-1
b=(4^101-1):3
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