1/

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

\(=\left(1.2+2.3+3.4+...+99.100\right)-\left(1+2+3+...+99\right)=\)

Đặt 

\(A=1.2+2.3+3.4+...+99.100\)

\(3A=1.2.3+2.3.3+3.4.3+...+99.100.3=\)

\(=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)=\)

\(=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-...-98.99.100+99.100.101=\)

\(=99.100.101\Rightarrow A=\dfrac{99.100.101}{3}=33.100.101\)

Đặt

\(B=1+2+3+...+99=\dfrac{99.\left(1+99\right)}{2}=4950\)

\(\Rightarrow N=A-B\)

2/

Số hạng cuối cùng là 10000 hoặc 1000000 mới làm được

\(A=1^2+2^2+3^2+...+100^2\) 

Tính như câu 1

3/ Làm như bài 4

4/

\(S=1^2+3^2+5^2+...+99^2=\)

\(=1.\left(3-2\right)+3\left(5-2\right)+5\left(7-2\right)+...+99\left(101-2\right)=\)

\(=\left(1.3+3.5+5.7+...+99.101\right)-2\left(1+3+5+...+99\right)\)

Đặt

\(B=1+3+5+...+99=\dfrac{50.\left(1+99\right)}{2}=2500\) 

Đặt

\(A=1.3+3.5+5.7+...+99.101\)

\(6A=1.3.6+3.5.6+3.7.6+...+99.101.6=\)

\(=1.3.\left(5+1\right)+3.5.\left(7-1\right)+5.7.\left(9-3\right)+...+99.101.\left(103-97\right)=\)

\(=1.3+1.3.5-1.3.5+3.5.7-3.5.7+5.7.9-...-97.99.101+99.101.103=\)

\(=3+99.101.103\Rightarrow A=\dfrac{3+99.101.103}{6}\)

\(\Rightarrow S=A-2B\)

Bài 1:

\(N=1^2+2^2+3^3+...+99^2\)

\(N=1.1+2.2+3.3+...+99.99\)

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

\(N=1.2-1+2.3-2+3.4-3+...+99.100-99\)

\(N=\left(1.2+2.3+3.4+...+99.100\right)-\left(1+2+3+...+99\right)\)

Đặt \(\left\{{}\begin{matrix}A=1.2+2.3+3.4+...+99.100\\B=1+2+3+...+99\end{matrix}\right.\)

+) Tính \(A=1.2+2.3+3.4+...+99.100\)

Ta có:

\(3A=1.2.3+2.3.3+3.4.3+...+99.100.3\)

\(3A=1.2.3+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100\)

\(3A=99.100.101\)

\(\Rightarrow A=\dfrac{99.100.101}{3}=333300\)

+) Tính \(B=1+2+3+...+99\)

\(B\) có số số hạng là: \(\dfrac{99-1}{1}\) + 1 = 99 (số hạng)

\(\Rightarrow B=\dfrac{\left(99+1\right).99}{2}=4950\)

\(\Rightarrow N=A-B=333300-4950=328350\)

\(\Rightarrow N=328350\)

xin loi mik danh nham nhe bai do la 10000 nhe

Cộng thêm 1/2 vào biểu thức đã cho, có:

S + 1/2= 1/2+1/4+ 1/8+ 1/16+1/32+1/64+1/128

Nhận xét:

BÀI 1:

\(S=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

\(S=1+\frac{1}{1.2}+\frac{1}{2.2}+\frac{1}{2.4}+\frac{1}{4.4}+\frac{1}{4.8}\)

\(S=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}\)

\(S=1+1-\frac{1}{8}\)

\(S=\frac{15}{8}\)

BÀI 2:

\(A=1.2+2.3+3.4+...+98.99\)

\(\Rightarrow3A=1.2.3+2.3.3+3.4.3+...+98.99.3\)

\(3A=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+98.99.\left(100-97\right)\)

\(3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+98.99.100-97.98.99\)

\(3A=\left(1.2.3+2.3.4+3.4.5+98.99.100\right)-\left(1.2.3+2.3.4+...+97.98.99\right)\)

\(3A=98.99.100\)

\(3A=970200\)

\(\Rightarrow A=970200:3\)

\(A=323400\)

CHÚC BN HỌC TỐT!!!
 

Xét TH1 : ( S < 8/9 )

\(\frac{1}{2\cdot2}< \frac{1}{1\cdot2};\frac{1}{3\cdot3}< \frac{1}{2\cdot3};...;\frac{1}{9\cdot9}< \frac{1}{8\cdot9}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\)

hay \(S< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}\)

\(S< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\)

\(S< 1-\frac{1}{9}=\frac{8}{9}\left(1\right)\)

TH2 : ( S > 2/5 )

\(\frac{1}{2\cdot2}>\frac{1}{2\cdot3};\frac{1}{3\cdot3}>\frac{1}{3\cdot4};...;\frac{1}{9\cdot9}>\frac{1}{9\cdot10}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{9^2}>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)

hay \(S>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)

\(S>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(S>\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\left(2\right)\)

Từ (1) và (2) => đpcm

Ko tk thì ko phải là ng` nx rồi :)

a)

Vì 2/9=6/27=8/36=12/54=16/72=18/81 nên:

2/9+6/27+8/36+12/54+16/72+18/81=

2/9+2/9+2/9+2/9+2/9+2/9=

2/9*6=

12/9=

4/3

Vậy 2/9+6/27+8/36+12/54+16/72+18/81=4/3

b)

Ta có:

1-2/5=3/5

1-2/7=5/7

1-2/9=7/9

...

1-2/99=97/99

Vậy (1-2/5)*(1-2/7)*(1-2/9)*...*(1-2/99)=

3/5*5/7*7/9*...*97/99=

(3*5*7*...*97)/(5*7*9*...*99)=

3/99=

1/33

Vậy (1-2/5)*(1-2/7)*(1-2/9)*...*(1-2/99)=1/33

c)

Gọi biểu thức 1/2+1/4+1/8+1/16+...+1/1024 là S,ta có:

S=1/2+1/4+1/8+1/16+...+1/1024

S*2=1+1/2+1/4+1/8+...+1/512

S*2-S=(1+1/2+1/4+1/8+...+1/512)-(1/2+1/4+1/8+1/16+...+1/1024)

S=1-1/1024

S=1023/1024

Vậy 1/2+1/4+1/8+1/16+...+1/1024=1023/1024

Cảm ơn bạn nhé!

giúp em với ạ 8:45 em đi học thêm rồi

S=1/2+1/4+1/8+1/16+1/32+1/64

S=1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32+1/32-1/64

S=1-1/64

S=63/64

S = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64

2S = 1/2 x 2 +1/4 x 2 + 1/8 x 2 + 1/16 x 2 + 1/32 x 2 + 1/64 x 2

2S =1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32

2S - S = ( 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 ) - ( 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 )

S = 1 - 1/64

S = 63/64

a, = 1

b 1 = S