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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
Ta có: (x1+x2)+(x3+x4)+...+(x99+x100)+x101=0 (50 nhóm)
=1x50+x101=0
=50 + x101=0
x101=0-50=-50
\(x+\left(x-1\right)+\left(x-2\right)+...+\left(x-101\right)=-516\)
\(x+x-1+x-2+...+x-101=-516\)
\(\left(x+x+...+x\right)-\left(1+2+...+101\right)=-516\)
\(102x-\left[\left(101+1\right)101:2\right]=-516\)
\(102x-5151=-516\)
\(102x=4635\)
\(x=\dfrac{1545}{34}\)
A = \(\dfrac{3^{100}.\left(-2\right)+3^{101}}{\left(-3\right)^{101}-3^{100}}\)
A = \(\dfrac{3^{100}.\left(-2\right)+3^{100}.3}{\left(-3\right)^{100}.\left(-3\right)-3^{100}}\)
A = \(\dfrac{3^{100}.\left(-2+3\right)}{3^{100}.\left(-3\right)-3^{100}}\)
A = \(\dfrac{3^{100}.1}{3^{100}.\left(-3-1\right)}\)
A = \(\dfrac{3^{100}}{3^{100}}\) . \(\dfrac{1}{-4}\)
A = - \(\dfrac{1}{4}\)
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
( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 101 ) = 9780
( x + x + x +... + x ) + ( 1 + 2 + 3 + ... + 101 ) = 9780
x * 101 + 5151 = 9780
x * 101 = 9780 - 5151
x * 101 = 4629
x = 4629 : 101
x = 45,83