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\(D=\left(2^9.3+2^9.5\right)-2^{12}\)
\(D=2^9.\left(3+2\right)-2^{12}\)
\(D=2^9.5-2^{12}\)
\(D=512.5-4096\)
\(D=2560-4096\)
\(D=-1536\)
\(\left(1+2+3+...+100\right).\left(1^2+2^2+3^2+...+100^2\right).\left(65.111-13.15.37\right)\)
\(=\left(1+2+3+...+100\right).\left(1^2+2^2+3^2+...+100^2\right).\left(7215-7215\right)\)
\(=\left(1+2+3+...+100\right).\left(1^2+2^2+3^2+...+100^2\right).0\)
\(=0\)
D=(29.3+29.5)-212
D=((29.(3+5))-212
D=(29.8)-212
D=(29.23)-212
D=29+3-212
D=212-212
D=0
(1+2+3+...+100).(12+22+32+....+1002).(65.111-13.15.37)
=(1+2+3+...+100).(12+22+32+....+1002).7215-7215
=(1+2+3+...+100).(12+22+32+....+1002).0
=0
C=210-2
C=29+1-2
C=29.2-2
C=2.(29-1)
C=2.(512-1)\
C=2.511
C=1022
F=1+31+32+33+......+3100
F=3+31+32+33+......+3100
3F=3.(3+31+32+33+......+3100)
3F=32+32+33+34+......+3100
3F-F=3100+32-3-3
2F=3100+9-3-3
F=\(\frac{3^{100}+3}{2}\)
Chúc bn học tốt
\(F=2+2^2+2^3+...+2^{100}\)
\(2F=2^2+2^3+2^4+...+2^{101}\)
\(2F-F=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+2^3+...+2^{100}\right)\)
\(F=2^{101}-2\)
Vậy...
\(E=3^0+3^1+3^2+...+3^{100}\)
\(E=1+3+3^2+...+3^{100}\)
\(3E=3+3^2+...+3^{101}\)
\(3E-3E=\left(3+3^2+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)
\(2E=3^{101}-1\)
\(E=\frac{3^{101}-1}{2}\)
Vậy...
100 - 99 + 98 - 97 ...... + 4 - 3 +2 - 1
= ( 100 - 99) + ( 98 - 97) + ( 96 - 95) +.....( 4 - 3) + ( 2 - 1)
= 1 + 1 + ....+1 + 1 ( 50 số 1 ) = 50
tick đúng cho mình nha ^_^
= (100 - 99 ) + ( 98 - + 97 ) + . . . + ( 4 - 3 ) + ( 2 - 1 )
= 1 + 1 + . . . + 1 + 1
50 số hạng
= 1 x 50 = 50
ta có
\(B=1+\left(1-\frac{1}{2}\right)+..+\left(1-\frac{1}{100}\right)\)
\(=1+\frac{1}{2}+\frac{2}{3}+..+\frac{99}{100}=A\)
Vậy A=B
Ta có : 13+ 23+...+1003 = (1+2+3+...+100)2
Đặt A = 1+2+3+...+100
⇒A= \(\dfrac{\left(100+1\right).100}{2}\) ⇔A= 5050
⇒50502 = (x-1)2 ⇔ x=5051
B=1+3+3^2+3^3+..+3^100
B=1+3+3^2+3^3+..+3^100
3B = 3 + 3^2 ... 3^101
3B - B = 2B = (3 + 3^2 ... 3^101) - (1+3+3^2+3^3+..+3^100)
B = (3^101 - 1) : 2
Đặt A=1+3+3^2+3^3+...+3^100
\(\Rightarrow\)3A=3+3^2+3^3+...+3^101
Ta có : 3A-A=(3+3^2+3^3+...+3^101)-(1+3+3^2+3^3+..+3^100)
2A=3+3^2+3^3+...+3^101-1-3-3^2-3^3-..-3^100
2A=3^101-1\(\Rightarrow\)\(A=\frac{3^{101}-1}{2}\)