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M=2/3.5+2/5.7+2/7.9+....+2/97.99
M=1.(1/3-1/5+1/5-1/7+...+1/97-1/99)
M=1.(1/3-1/99)
M=32/99
0 + 1 + 2 + .... + 98 + 99 + 100
= ( 100 + 1 ) + ( 99 + 2 ) + ( 98 + 3 ) + .... + ( 3 + 98 ) + ( 2 + 99 ) + ( 1 + 100 )
= 101 + 101 + 101 + .... + 101 + 101 + 101 ( có 50 số hạng 101 )
= 101 . 50
= 5050
#)Giải :
\(A=1+2+2^2+...+2^{100}\)
\(2A=2+2^2+2^3+...+2^{101}\)
\(2A-A=\left(2+2^2+2^3+...+2^{101}\right)-\left(1+2+2^2+...+2^{100}\right)\)
\(A=2^{101}-1\)
\(B=1+3^2+3^4+...+3^{100}\)
\(3^2B=3^2+3^4+3^6+...+3^{102}\)
\(3^2B-B=\left(3^2+3^4+3^6+...+3^{102}\right)-\left(1+3^2+3^4+...+3^{100}\right)\)
\(8B=3^{102}-1\)
\(B=\frac{3^{102}-1}{8}\)
\(C=1+5^3+5^6+...+5^{99}\)
\(5^2C=5^3+5^6+5^9+...+5^{102}\)
\(5^2C-C=\left(5^3+5^6+5^9...+5^{102}\right)-\left(1+5^3+5^6+...+5^{99}\right)\)
\(24C=5^{102}-1\)
\(C=\frac{5^{102}-1}{24}\)
a) A = 1 + 22 + ... + 2100
=> 2A = 22 + 23 + ... + 2101
Lấy 2A - A = (2 + 22 + ... + 2101) - (1 + 22 + ... 2100)
A = 2101 - 1
b) B = 1 + 32 + 34 + ... + 3100
=> 32B = 32 + 34 + 36 + ..... + 3102
=> 9B = 32 + 34 + 36 + ..... + 3102
Lấy 9B - B = ( 32 + 34 + 36 + ..... + 3102) - (1 + 32 + 34 + ... + 3100)
8B = 3102 - 1
B = \(\frac{3^{102}-1}{8}\)
c) C = 1 + 53 + 56 + ... + 599
=> 53.C = 53 . 56 . 59 + ... + 5102
=> 125.C = 53 . 56 . 59 + ... + 5102
Lấy 125.C - C = (53 . 56 . 59 + ... + 5102) - (1 + 53 + 56 + ... + 599)
124.C = 5102 - 1
=> C = \(\frac{5^{102}-1}{124}\)
a, => 3x-17 = 0 hoặc 3x-17 = 1
=> x=17/3 hoặc x=6
b, => x+1+x+2+....+x+100=205550
=>100x + (1+2+...+100)=205550
=> 100x + 5050 = 205550
=> 100x = 205550 - 5050 = 200500
=>x= 2005
c,=>x+x+1+....+x+2010=2029099
=>2011x+(1+2+....+2010)=2029099
=>2011x+2021055=2029099
=>2011x = 2029099-2021055 = 8044
=>x=4
Có : 3Q = 3+3^2+....+3^101
2Q=3Q-Q= (3+3^2+....+3^101)-(1+3+3^2+...+3^100) = 3^101-1
=>Q = (3^101-1)/2
cau 2
n+2=(n-3)+5
để n+2chia het cho n-3 thi 5 chia het cho n-3
=>n-3 la uoc cua 5
=>n-3 thuoc [-1;1;-5;5]
n-3=-1=>n=-1+3=2 n-3=1=>n=1+3=4
n-3=-5=>n=-5+3=-2 n-3=5=>n=5+3=8
vay voi n=2 ;n=-2;n=4;n=8 thi n+2 chia het cho n-3
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=2-\frac{2}{3}+\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{99}-\frac{2}{101}\)
\(=2-\frac{2}{101}=\frac{200}{101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-.....+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)