2² + 4² + 6² + ... + 16² + 18²

= 4.(1 + 2² + 3² + ... + 8² + 9²)

= 4.285

= 1140

= 285 nha mình ghi nhầm thành 385

q=2²+4²+6²+...+18²

q= 1².2²+2².2²+3².2²+...+9².2²

q= 2².(1²+2²+3²+...+9²)

q= 2².285=4.285=1140

Ta có \(2^2+4^2+...+20^2=2^2\left(1^2+2^2+...+10^2\right)=2^2.385=1540\).

\(B=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

\(B=\dfrac{1}{2.2}+\dfrac{1}{4.4}+...+\dfrac{1}{100.100}\)

\(B=\dfrac{1}{2}-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+...+\dfrac{1}{100}-\dfrac{1}{100}\)

\(B=0+0+...+0\)

\(B=0\)

B=221​+421​+621​+...+10021​

�=12.2+14.4+...+1100.100B=2.21​+4.41​+...+100.1001​

�=12−12+14−14+...+1100−1100B=21​−21​+41​−41​+...+1001​−1001​

$B=0+0+...+0$

$B=0$

$tick\; cái$

bạn sao chép của người bạn Ninh mà

S = 22 + 42 + 62 + ... + 202

   = (2.1)2 + (2.2)2 + (2.3)2 ... (2.10)2

   = 22.12 + 22.22 + 22.32 + ... + 22.102

   = 22 (12 + 22 + ... + 102 )

   = 4 . 385 = 1540

Ta có : \(1^2+2^2+3^2+.....+10^2=385\)

\(\Leftrightarrow2^2\left(1^2+2^2+3^2+.....+10^2\right)=2^2.385\)

\(\Leftrightarrow2^2+4^2+6^2+.....+20^2=4.385\)

\(\Leftrightarrow2^2+4^2+6^2+.....+20^2=1540\)

Sửa đề: CHo 1²+2²+...+10²=385. Tính S = 2²+4² +...+ 20²

S = 2² + 4² +...+ 20²

= (1.2)² + (2.2)² +...+ (2.10)²

= 1².2² + 2².2² +...+ 2².10²

= 2²(1² + 2² +...+ 10²)

= 4.385

= 1540

S= 1^2  .  2^2 + 2^2 . 2^2  + ... + 10^2 . 2^2
  = 4. ( 1^2 + 2^2 +...+ 10^2)  = 4 . 385 = 1540

chúc bạn học giỏi

= 50

Nho t ick nha

     \(\left(102+82+62+42+22\right)-\left(12+32+53+72+92\right)\)

\(=102+82+62+42+22-12-32-52-72-92\)

\(=\left(102-92\right)+\left(82-72\right)+\left(62-52\right)+\left(42-32\right)+\left(22-12\right)\)

\(=10+10+10+10+10\)

\(=10.5\)

\(=50\)

Giải:

a) Ta có:

1/2²=1/2.2 < 1/1.2

1/3²=1/3.3 < 1/2.3

1/4²=1/4.4 < 1/3.4

1/5²=1/5.5 < 1/4.5

1/6²=1/6.6 < 1/5.6

1/7²=1/7.7 < 1/6.7

1/8²=1/8.8 <1/7.8

⇒B<1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8

   B<1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8

   B<1/1-1/8

   B<7/8

mà 7/8<1

⇒B<7/8<1

⇒B<1

b)S=3/1.4+3/4.7+3/7.10+...+3/40.43+3/43.46

   S=1/1-1/4+1/4-1/7+1/7-1/10+...+1/40-1/43+1/43-1/46

   S=1/1-1/46

   S=45/46

Vì 45/46<1 nên S<1

Vậy S<1

Chúc bạn học tốt!

a)\(\dfrac{1}{2^2}<\dfrac{1}{1.2}\)

\(\dfrac{1}{3^3}<\dfrac{1}{2.3}\)

\(...\)

\(\dfrac{1}{8^2}<\dfrac{1}{7.8}\)

Vậy ta có biểu thức:

\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{8^2}<\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{7.8}\)

\(B= 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{7}-\dfrac{1}{8}\)

\(B<1-\dfrac{1}{8}=\dfrac{7}{8}<1\)

Vậy B < 1 (đpcm)

Lời giải:

Gọi vế trái là $A$

$2A=\frac{2}{2^2}+\frac{2}{4^2}+\frac{2}{6^2}+...+\frac{2}{2022^2}$

Xét số hạng tổng quát:

$\frac{2}{n^2}$. Ta sẽ cm $\frac{2}{n^2}< \frac{1}{(n-1)n}+\frac{1}{n(n+1)}(*)$

$\Leftrightarrow \frac{2}{n^2}< \frac{n+1+n-1}{n(n-1)(n+1)}$

$\Leftrightarrow \frac{2}{n^2}< \frac{2}{(n-1)(n+1)}$

$\Leftrightarrow \frac{2}{n^2}< \frac{2}{n^2-1}$ (luôn đúng)

Thay $n=2,4,...., 2022$ vào $(*)$ ta có:

$\frac{2}{2^2}< \frac{1}{1.2}+\frac{1}{2.3}$

$\frac{2}{4^2}< \frac{1}{3.4}+\frac{1}{4.5}$

.......

Suy ra: $2A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2022.2023}$

$2A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2022}-\frac{1}{2023}$

$2A< 1-\frac{1}{2023}< 1$

$\Rightarrow A< \frac{1}{2}$

ta có: 12 + 62 + 82=1 + 36 + 64 = 101

22 + 42 + 92 = 4 + 16 + 81 = 101

Vậy 12 + 62 + 82 = 22 + 42 + 92