3/5+3/15+3/35+3/63+............+3/nx(n+2)=3015/2011
3/5+3/15+3/35+3/63+............+3/nx(n+2)=3015/2011
3/5+3/15+3/35+3/63+......+3/nx(n+2)=3015/2011
Find n such that: 2/3+2/15+2/35+...+ 2/ nx(n+2)=322/323
2/3+2/15+2/35+...+2/n x (n+2)=322/323
=2/1.3+2.3.5+2/5.7+...+2/n x(n+2)=322/323
= n+2 =323
n=323-2
n=321
2/3+2/15+2/35+...+2/n x (n+2) = 322/323
2/1x3+2/3x5+2/5x7 +...+ 2/nx(n+2) = 322/323
1-1/3+1/3-1/5+1/5-1/7+...+1/n-1/(n+2) = 322/323
1-1/n+2 = 322/323
1/n+2 = 1-322/323
1/n+2 = 1/323
=> n+2 = 323
n = 323 - 2 = 321
a/ 1/2 + 5/6 + 11/12 + 19/20
b/ 1/2 + 5/6 + 11/12 + 19/20 + 29/30 + 41/42
c/ (1-1/3) + (1-1/15) + (1-1/35) + (1-1/63)
d/ 1/2 + 5/6 + 11/12 + ... + 9899/9900
e/ 2/3 + 14/15 + 34/35 +62/63
f/ 2/3 + 14/15 + 34/35 + ... + 9998/9999
cái này tính cái gì thế
ko hiểu
a) A = 1/15+1/35+1/63+1/99+1/143+1/195
a) M= 1+3+3^2+3^3+...+3^25 và N = 3^26 : 2 Tính N-M
a)\(A=\frac{1}{15}+\frac{1}{35}+...+\frac{1}{195}=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{13.15}=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{2}{15}\)
b)\(M=1+3+3^2+...+3^{25}=\frac{3^{26}-1}{3-1}=\frac{3^{26}-1}{2}
Tính :
a) ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 )
b) \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
a) ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 )
vì ( 125125 x 127 - 127127 x 125 ) =[125125 x (125+2)] - 127127 x 125 ) =>125125 x (125+2)=125.125125+125125.2=125125.125+250250=125125.125+125.2002=125.(125125+2002)=125.127127
=> ( 125125 x 127 - 127127 x 125 )=127127.125-127127.125=0
=> (1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 ) =0
a) ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x ( 125125 x 127 - 127127 x 125 )
= ( 1 + 3 + 5 + 7 + ....... + 2007 + 2009 + 2011 ) x 0
= 0
b, \(\frac{1}{3}\)+ \(\frac{1}{15}\)+ \(\frac{1}{35}\)+ \(\frac{1}{63}\)+ \(\frac{1}{99}\)+ \(\frac{1}{143}\)+ \(\frac{1}{195}\)
= \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{5}\)+ \(\frac{1}{5}\)- \(\frac{1}{7}\)+\(\frac{1}{7}\)- \(\frac{1}{9}\)+...........+\(\frac{1}{13}\)- \(\frac{1}{15}\)
= \(\frac{1}{3}\)- \(\frac{1}{15}\)
= \(\frac{4}{15}\)
5-2/3-14/15+1/35-62/63-98/99-142/143
\(5-\dfrac{2}{3}-\dfrac{14}{15}+\dfrac{1}{35}-\dfrac{62}{63}-\dfrac{98}{99}-\dfrac{142}{143}\)
\(=5-\left(1-\dfrac{1}{3}\right)-\left(1-\dfrac{1}{15}\right)+\dfrac{1}{35}-\left(1-\dfrac{1}{63}\right)-\left(1-\dfrac{1}{99}\right)-\left(1-\dfrac{1}{143}\right)\)
\(=5-1+\dfrac{1}{1\cdot3}-1+\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}-1+\dfrac{1}{7\cdot9}-1+\dfrac{1}{9\cdot11}-1+\dfrac{1}{11\cdot13}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{11}-\dfrac{1}{13}\)
\(=1-\dfrac{1}{13}=\dfrac{12}{13}\)
tìm số tự nhiên n biết : 2/3+2/15+2/35+2/63+...+2/n = 100/101
=> 2/1x3 +2/3x5+2/5x7+2/7x9+...+2/nx(n+2)
=>1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9+...+1/n-1/n+2
=>1-1/n+2=100/101
1/n+2=1-100/101
1/n+2=1/101
=>n+2=101
=>n=101-2
=>n=99
Sử dụng tính chất a.b +(-) a.c = a.(b+c)
1. (15 . 13 - 21) : 2 + 108
2. 4^2 . 444446 - 4^3 . 111111
3. (5^3015 - 5^3013) : 5^3013
4. 125 - [ 65 - (7 - 3)^3 ]
5. 123 . 456456 - 228. 246246
6. (548 . 13 + 1096) . 43 : 15 . 86
7. (3^13 : 99 - 15 . 3^14) : (3^17 : 3^2)
Cảm ơn !
1: =(195-21):2+108=174:2+108=87+108=195
2: \(=16\left(444446-444444\right)=16\cdot2=32\)
3: \(=\dfrac{5^{3015}}{5^{3013}}-\dfrac{5^{3013}}{5^{3013}}=25-1=24\)
4: \(=125-\left[65-4^3\right]=125-1=124\)
6: \(=8210\cdot43:15\cdot86=2024038.667\)