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12+22+32+42+....+122=650
=>(1+2+3+4+...+12)2=650
=>((1+2+3+4...+12)2)x2=650x2
=>(2+4+6+...12+24)2=1300
=>22+42+62+....+242=1300
Có : \(2^2+4^2+6^2+...+12^2+24^2\)
=\(1^2.2^2+2^2.2^2+3^2.2^2+4^2.2^2+...+12^2.2^2\)
=\(2^2.\left(1^2+2^2+3^2+4^2+...+12^2\right)\)
Theo đề bài ta có : \(2^2+4^2+6^2+...+12^2+24^2=2^2.650\)
=2600
Ta có:
\(1^2\)+\(2^2\)+\(3^2\)+...+\(12^2\)=650
=> \(2^2\).(\(1^2\)+\(2^2\)+\(3^2\)+...+\(12^2\))=\(2^2\).650
\(2^2\)+\(4^2\)+\(6^2\)+....+\(24^2\)=2600
Vậy \(2^2\)+\(4^2\)+\(6^2\)+....+\(24^2\)=2600.
S = 2^2.1 + 2^2.2^2 + 3^2.2^2+....+2^2.10^2
S = 2^2( 1 + 2^2 + 3^2 +.. +10^2)
s = 2^2.385
s = 4.385
S = 1540
Đúng cho mình nha
a: \(=\left(\dfrac{5}{15}-\dfrac{12}{9}\right)+\left(\dfrac{14}{15}+\dfrac{11}{25}\right)+\dfrac{2}{7}\)
\(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{70+33}{75}+\dfrac{2}{7}\)
\(=-1+\dfrac{2}{7}+\dfrac{103}{75}=\dfrac{-5}{7}+\dfrac{103}{75}=\dfrac{346}{525}\)
b: \(4\cdot\left(-\dfrac{1}{2}\right)^3+\dfrac{1}{2}\)
\(=4\cdot\dfrac{-1}{8}+\dfrac{1}{2}=\dfrac{-1}{2}+\dfrac{1}{2}=0\)
c: \(\dfrac{10^3+5\cdot10^2+5^3}{6^3+3\cdot6^2+3^3}=\dfrac{5^3\cdot8+5\cdot5^2\cdot2^2+5^3}{3^3\cdot2^3+3\cdot2^2\cdot3^2+3^3}\)
\(=\dfrac{5^3\left(8+4+1\right)}{3^3\left(8+4+1\right)}=\dfrac{125}{27}\)
e: \(\dfrac{2^8\cdot9^2}{6^4\cdot8^2}=\dfrac{2^8\cdot3^4}{3^4\cdot2^4\cdot2^6}=\dfrac{1}{4}\)
22 + 42 + 62 + ... + 242
= 22 . ( 12 + 22 + 32 + ... + 122 )
= 22 . 650
= 2600
Ta có : 22 + 42 + 62 + ... + 242
= 12 . 22 + 22 . 22 + 32 . 22 + .... + 122 . 22
= (12 + 22 + 32 + ... + 122) . 22
= 650 . 4 = 2600