B = 7+7^2+7^3+...+7^99
Những câu hỏi liên quan
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
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!
a, A= 1+4+4^2+4^3+...+4^100
b, B=7+7^3+7^5+...+7^99
c, C=2+2^3+2^5+2^7+...+2^2009
d, N= 1+2^2+3^2+4^2+5^2+...+99^2
MOI NGUOI GIUP MINH VOI
Tính:
a,B=1+6^2+6^4+...+6^2016
b,C=2+2^3+2^5+...+2^2017
c,D=1-7+7^2-7^3+...-7^99+7^100
d,E=100^2-99^2+98^2_97^2+...+2^2-1^2
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 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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tính nhanh
B=7^101-7^100-7^99-7^98 -... - 7^2 - 7-1.
C=1-3+3^2-3^3+...+3^48-3^49+3^50
B = 7101-7100-799-...-7-1
B = -(7101+7100+799+...+7+1)
Đặt D = 1+7+72+....+7101
7D = 7+72+73+...+7102
6D = 7D - D = 7102-1
=> D = \(\frac{7^{102}-1}{6}\)
=> B = \(-\left(\frac{7^{102}-1}{6}\right)\)
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Tớ chắc là cậu chép sai đề
Tớ thấy nó hơi vô lí
Chúc bạn hok tốt!
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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, \(E=\frac{2^{9011}-2}{3}\)
A = 1 + 3 mũ 2 + 3 mũ 4 + 3 mũ 6 + 3 mũ 8 +...+ 3 mũ 100
B = 7 + 7 mũ 3 + 7 mũ 5 + 7 mũ 7 + ....+ 7 mũ 99
A = 1 + 32+34+...+3100
9A= 32+34+36+...+3102
=>9A-A=(32+34+36+...+3102)-(1 + 32+34+...+3100)
<=>8A=3102-1
=>A=\(\frac{3^{102}-1}{8}\)
Tương tự với câu B, nhân B cho 72=49
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B=7+73+75+...+799
49B=73+75+77+...+7101
49B-B=7101-7
=>B=\(\frac{7^{101}-7}{48}\)
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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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