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(x+1)+(x+3)+(x+3^2)+...+(x+3^99)+(x+3^100)=3^101
=>x+1+x+3+x+3^2+...+x+3^99+x+3^100=3^101
=>(x+x+...+x) + (1+3+3^2+...+3^100) = 3^101
101 lần
=>101.x + (1+3+3^2+...+3^100) = 3^101.
Đặt A = 1 + 3 + 3^2 + ... + 3^100
=> 3A = 3.(1 + 3 + 3^2 + ... + 3^100)
=> 3A = 3 + 3^2 + 3^3 + ... + 3^101
=> 3A - A = 3 + 3^2 + 3^3 + ... + 3^101 - 1 - 3 - 3^2 - ... - 3^100
=> 2A = 3^101 - 1
=> A = (3^101 - 1)/2
Thay A = (3^101 - 1)/2 vào trên, ta có:
101.x + (3^101 - 1)/2 = 3^101
=>101x = 3^101 - (3^101-1)/2
=> x = [3^101 - (3^101 - 1)/2]/101
a: S=1(1+1)+2(1+2)+...+100(1+100)
=1+2+...+100+1^2+2^2+...+100^2
\(=\dfrac{100\cdot102}{2}+\dfrac{100\cdot\left(100+1\right)\cdot\left(2\cdot100+1\right)}{6}\)
\(=100\cdot51+\dfrac{100\cdot101\cdot201}{6}\)
=343450
b: \(A=1\cdot2\cdot3+2\cdot3\cdot4+...+100\cdot101\cdot102\)
=>\(4\cdot A=1\cdot2\cdot3\cdot\left(4-0\right)+2\cdot3\cdot4\left(5-1\right)+...+100\cdot101\cdot102\left(103-99\right)\)
=>4*A=100*101*102*103
=>A=25*101*102*103
S= 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100
S x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3
S x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)
S x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.
S x 3 = 99x100x101 A = 99x100x101 : 3 A = 333300
Tính x1 + x2 +...+ x99 + x100 + x101 = 0
(x1 + x2)+ ...+ ( x99 + x100)+ x101 = 0
1 + ... + 1 + x101 = 0
1 x 50 + x101 = 0
50 + x101 = 0
x101 = 0 - 50
x101 = -50
Ta có: x100 + x101 = 1
x100 + (-50) = 1
x100 = 1-(-50)
x100 =51
Vậy x101 = 51
1
b;
B=1+ (7-5) + (11-9) + ...+(101-99)
B=1+2+2+..+2
B=1+25.2=51
2.
a.
ĐK : x+2 >=0 => x>=-2
\(\left|x+2\right|-x=2\\ \Rightarrow\left|x+2\right|=2+x\\ \Rightarrow\left[{}\begin{matrix}x+2=x+2\\x+2=-x-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}0x=0\\2x=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}0x=0\\x=-2\end{matrix}\right.\)
Vậy x=-2