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A = \(2^2.\left(1^2+2^2+3^2+...+10^2\right)=4.385=1540\)
B=\(3^2.\left(1^2+2^2+3^2+...+10^2\right)=385.9=3465\)
Bài 1:
A = 1 + 3 + 32 + ... + 3100
=> 3A = 3 + 32 + ... + 3101
=> 2A = 3101 - 1
=> A = \(\frac{3^{101}-1}{2}\)
B = 1 + 42 + 44 + ... + 4100
=> 8B = 42 + 44 + ... + 4102
=> 7B = 4102 - 1
=> B = \(\frac{4^{102}-1}{7}\)
Bài 2:
a) S1 = 22 + 42 + ... + 202
=> S1 = 22(1+22+...+102)
=> S1 = 22.385
=> S1 = 1540
b) S2 = 1002 + 2002 + ... + 10002
=> S2 = 1002(1+22+...+102)
=> S2 = 1002.385
=> S2 = 3850000
S=22+42+62+...+202
S=(2*1)2+(2*2)2+(2*3)2+...+(2*10)2
=22*12+22*22+...+22*102=22(12+22+32+...+102)=4*385=1540
Vậy S=1540
a: A=3^2(1^2+2^2+...+10^2)
=9*385
=3465
b: B=2^3(1^3+2^3+...+10^3)
=8*3025
=24200
Lời giải:
$S=10^2+(10.2)^2+(10.3)^2+...+(10.9)^2+(10.10)^2$
$=10^2(1^2+2^2+3^2+...+9^2+10^2)$
$=100.385=38500$
Ta có:
\(2^2\left(1^2+2^2+3^2+...+10^2\right)=2^2+4^2+6^2+...+20^2=S\)
=> \(S=2^2.385=1540\)
S = 2^2 + 4^2+6^2+.....+20^2
= ( 1.2 ) ^2 + ( 2.2)^2 +.....+ (2.10 ) ^2
= 2^2( 1^2 + 2^2 +.....+ 10^2 )
=2^2 . 385
= 4 . 385 = 1540
Bài 1:
\(S=2^2+4^2+6^2+...+20^2\)
\(=\left(1\cdot2\right)^2+\left(2\cdot2\right)^2+\left(2\cdot3\right)^2+...+\left(2\cdot10\right)^2\)
\(=1\cdot2^2+2^2\cdot2^2+2^2\cdot3^2+...+2^2\cdot10^2\)
\(=2^2\left(1+2^2+3^2+...+10^2\right)\)
\(=4\cdot385=1540\)
Bài 2:
\(A=2^0+2^1+2^2+...+2^{100}\)
\(A=1+2+2^2+...+2^{100}\)
\(2A=2\left(1+2+2^2+...+2^{100}\right)\)
\(2A=2+2^2+2^3+...+2^{101}\)
\(2A=\left(2+2^2+...+2^{101}\right)-\left(1+2+...+2^{100}\right)\)
\(A=2^{101}-1\)
Giải:
\(1.\) \(S=2^2+4^2+6^2+....+20^2\)
\(2^2=\left(1.2\right)^2\)
\(4^2=\left(2.2\right)^2\)
\(...\)
Vế dưới \(= \left(1.2\right)^2 + \left(2.2\right)^2 + ...+ \left(9.2\right)^2+ \left(10.2\right)^2\)
\(= 2^2.(1^2 + 2^2 + 3^2 + ...+ 9^2 + 10^2) \)
\(= 4. 385\)
\(= 1540\)
\(2.\)
\( 2A = 2^1 + 2^2 + 2^3 + 2^4 +...+\)\(2^{2011}\)
\(2A - A = ( 2^1 + 2^2 + 2^3+ 2^4 +...+ 2^{2011} ) - ( 1 + 2^2 + 2^3 +...+ 2^{2010} ) \)
\(\Rightarrow A = 2^{2011} - 1\)