Bài 1. Tính các tổng sau:

1. S= 1+2+3+4+.................+98+99+100

S=( 100 - 1 ): 1 + 1 = 100

2. S= 2+4+6+8+.................+996+998

S = ( 998 - 2 ) : 2 + 1 = 499

3. S= 1.2+2.3+3.4+.............+98.99+99.100

S= 1.2 3-0 +2.3 (4-1) +3.4 

4. S= 1.2.3+2.3.4+3.4.5+..............+97.98.99+98.99.100

S= (100 -1) + 1 : 1 = 100

5. S= 1+2+3+..........+98+99+100

S=( 100 - 1) + 1   : 1

S= 100 

1.S=(1+100)+(2+99)+...(50+51)  (Tổng cộng có 50 cặp)

S=101+101+101+...101

S=101 x 50=5050

=>S= 5050

Câu a)
\(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(=\left(2^{100}+2^{99}+2^{98}+2^{97}+...+2^2+2\right)-2\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)\)
\(=\left(2^{100}+2^{99}+2^{98}+2^{97}+...+2^2+2\right)-\left(2^{100}+2^{98}+2^{96}+...+2^4+2^2\right)\)
\(=2^{99}+2^{97}+2^{95}+...+2^3+2\)
\(=\frac{2^2\cdot\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)-\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)}{3}\)
\(=\frac{\left(2^{101}+2^{99}+2^{97}+...+2^5+2^3\right)-\left(2^{99}+2^{97}+2^{95}+...+2^3+2\right)}{3}\)
\(=\frac{2^{101}-2}{3}\)

\(2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{2015.2016.2017}\)

\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{2.4}+...+\frac{1}{2015.2016}-\frac{1}{2016.2017}\)

\(2B=\frac{1}{1.2}-\frac{1}{2016.2017}\)

\(B=\frac{\frac{1}{1.2}-\frac{1}{2016.1017}}{2}\)

a. Áp dụng CT: n.9n+1)/2

=>S=(101.100)/2

b. SSH=(998-2) : 2+1

TBC=(998+2):2

Nhân SSH với TBC => S

c.

Đặt A= 1.2 + 2.3 + 3.4 + ...+ 99.100
 3A = 1.2.3+2.3.3+3.4.3+...+98.99.3+99.100.3
3A= 1.2.3+2.3(4-1)+3.4(5-2)+...+98.99(100-97)+99.100(101-98)
3A= 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...-97.98.99+99.100.101-98.99.100
3A = 99.100.101  3S = 3.33.100.101 
 A=33.100.101= 333300

d.

Đặt A=1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100

4A=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100)4

4A=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)

4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100

4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98.99.100.101

4A=98.99.100.101

=>A=98.99.100.101/4

a. S= 1+2+3+4+.....+98+99+100

S= (100 -1) : 1 + 1 =100

b. S= 2+4+6+8+.....+996+998

S= (998 -  2 ) : 2 + 1 = 499

c. S= 1.2+2.3+3.4+.....+98.99+99.100

Bài này hôm qua đã làm -.- vào thống kê của tôi mà nhìn :)

d. S= 1.2.3+2.3.4+3.4.5+......+97.98.99+98.99.100

S = (1.2.3.2.3.4.5.4.5.6+98.99.100)4

S=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+97.98.99+98.99.100

S=101 - 97

S=1.2.3.5.2.4.+2.1.2.3.4.3.4.5.5.6-2.4.5.4.5.6.7-3.4.5.6-3.4.5.6+.......100

S=1.2.3.3.4.5.5.6.7.7.8.9......+97.98.99+98.99.100

S=1.2.3.4.4.3.2.1+2.3.5-2.3.4.5+3.4.5.6.6.7.3.4.5.6+........97.98.99+98.99.100

S= 98.99.100.101

S=98.99.100.\(\frac{101}{4}\)

e. S= 1²+2²+3²+.....98²+99²+100²

S= 100² - 99² + 98² -97^{2 +...+} 2^2- 1²

S= (100 - 99) (100+99) (98 - 97) (98+97) +....+(2-1) (2+1)

S=(1+100) 100 :2

s=5050

Ngu như con bò

vay sao chi

Bài 1:

Ta thấy : \(\left\{\begin{matrix}\left(x-3\right)^2\ge0\\\left|y+1\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x-3\right)^2+\left|y+1\right|\ge0\)

\(\Rightarrow\left(x-3\right)^2+\left|y+1\right|-3\ge-3\)

\(\Rightarrow A\ge-3\)

Dấu "=" xảy ra khi \(\left\{\begin{matrix}\left(x-3\right)^2=0\\\left|y+1\right|=0\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}x-3=0\\y+1=0\end{matrix}\right.\)\(\Rightarrow\left\{\begin{matrix}x=3\\y=-1\end{matrix}\right.\)

Vậy \(Min_A=-3\) khi \(\left\{\begin{matrix}x=3\\y=-1\end{matrix}\right.\)

Bài 2:

\(S=1\cdot2\cdot3+2\cdot3\cdot4+...+97\cdot98\cdot99\)

\(4S=4\left(1\cdot2\cdot3+2\cdot3\cdot4+...+97\cdot98\cdot99\right)\)

\(4S=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot\left(5-1\right)+...+97\cdot98\cdot99\left(100-96\right)\)

\(4S=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4+...+97\cdot98\cdot99\cdot100-96\cdot97\cdot98\cdot99\)

\(4S=97\cdot98\cdot99\cdot100\Rightarrow S=\frac{97\cdot98\cdot99\cdot100}{4}=23527350\)

Bài 1:

$A=1.2+2.3+3.4+...+201.202$

$3A=1.2.3+2.3(4-1)+3.4(5-2)+....+201.202(203-200)$

$=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+201.202.203-200.201.202$

$=(1.2.3+2.3.4+3.4.5+...+201.202.203)-(1.2.3+2.3.4+....+200.201.202)$

$=201.202.203$

$\Rightarrow A=\frac{201.202.203}{3}=2747402$

Bài 2:

$S=4.5+5.6+6.7+....+100.101$

$3S=4.5(6-3)+5.6.(7-4)+6.7.(8-5)+....+100.101(102-99)$

$=4.5.6-3.4.5+5.6.7-4.5.6+6.7.8-5.6.7+....+100.101.102-99.100.101$

$=(4.5.6+5.6.7+6.7.8+...+100.101.102)-(3.4.5+4.5.6+5.6.7+...+99.100.101)$

$=100.101.102-3.4.5$

$\Rightarrow S=\frac{100.101.102-3.4.5}{3}=343380$

\(S=\left(2+4+...+100\right)-\left(1+3+5+...+99\right)\)

\(S=2+4+6+...+100-1-3-5-...-99\)

\(S=\left(2-1\right)+\left(4-3\right)+...+\left(100-99\right)\)

\(S=1+1+1+...+1\)

             50 số hạng

\(S=5.1=5\)

Cái cuối cho mình sửa lại thành S = 50.1 = 50 nha

S=(1-2)+(3-4)+(5-6)+...+(199-200)

S=(-1)+(-1)+...+(-1)

S=(-1).100=-100

S=1+(2-3)+(-4+5)+...+(98-99)+(-100+101)

S=1+(-1)+1+..+(-1)+1

S=1+25.(-1)+25.1

S=1+(-25)+25

S=1+0

=1

Nhanh nha mk cần gấp đó!