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2A=2101-2100+299-298+...............+23-22
2A+A=2101-2
A=(2101-2):3
Áp dụng công thức: (n-2)n(n+2) = n3 - 4n => n3 = (n-2).n.(n+2) + 4n
b18) Áp dụng: ta có: 23 = 4.2; 43 = 2.4.6 + 4.4 ; 63 = 4.6.8 + 4.6; ...; 1003 = 98.100.102 + 4.100
=> A = 4.2 + 2.4.6 + 4.4 + 4.6.8 + 4.6 +...+ 98.100.102 + 4.100
= (2.4.6 + 4.6.8 + 6.8.10 +....+ 98.100.102 ) + 4.(2 + 4 + 6 + ...+ 100) = B + 4.C
Tính B = 2.4.6 + 4.6.8 + 6.8.10 +....+ 98.100.102
=> 8.B = 2.4.6.8 + 4.6.8.8 + 6.8.10.8 +...+ 98.100.102.8
= 2.4.6.8 + 4.6.8 (10 - 2) + 6.8.10.(12 - 4) +...+ 98.100.102.(104 - 96)
= 2.4.6.8 + 4.6.8.10 - 2.4.6.8 + 6.8.10.12 - 4.6.8.10 +...+ 98.100.102.104 - 96.98.100.102
= (2.4.6.8 + 4.6.8.10 + 6.8.10.12 +...+ 98.100.102.104) - (2.4.6.8 + 4.6.8.10 +...+ 96.98.100.102)
= 98.100.102.104
=> B =98.100.102.104 : 8 = 12 994 800
C = 2+ 4+ 6 +..+100 = (2+100) . 50 : 2 = 2550
Vậy A = B +4C = 12 994 800 + 4. 2550 = 13 005 000
\(A=100^2-99^2+...+2^2-1^2\)
\(A=\left(100^2-99^2\right)+...+\left(2^2-1^2\right)\)
\(A=\left(100+99\right)\left(100-99\right)+...+\left(2-1\right)\left(2+1\right)\)
\(A=199+...+3\)
\(A+1+2=199+...+3+2+1\)
\(A+3=\frac{n\left(n+1\right)}{2}=\frac{199\left(199+1\right)}{2}\)
\(A+3=19900\Rightarrow A=19897\)
Bài 1 :
A=2+22+23+...+299+2100A=2+22+23+...+299+2100
⇒2A=22+23+24+...+2100+2101⇒2A=22+23+24+...+2100+2101
⇒A=2101−2⇒A=2101−2
B=3+32+33+...+399+3100B=3+32+33+...+399+3100
⇒3B=32+33+34+...+3100+3101⇒3B=32+33+34+...+3100+3101
Bài 2 :
2.Chứng minh rằng
212+312+213+214+315 chia hết cho 7
⇒2B=3101−3⇒2B=3101−3
⇒B=3101−32
P/s: làm từng phần một
1.
\(2A=2^2+2^3+...+2^{101}\)
\(2A-A=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)
\(A=2^{101}-2\)
2.
\(\frac{A}{2}=\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{59\cdot61}\)
\(\frac{A}{2}=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\)
\(\frac{A}{2}=\frac{1}{5}-\frac{1}{61}\)
\(\frac{A}{2}=\frac{56}{305}\)
\(A=\frac{112}{305}\)
A = 2100- 299 + 298 - 297 + ... + 22 - 2
=> 2A = 2101 - 2100 + 299 - 298 + ... + 23 - 22
Khi đó 2A + A = (2101 - 2100 + 299 - 298 + ... + 23 - 22) + (2100- 299 + 298 - 297 + ... + 22 - 2)
=> 3A = 2101 - 2
=> \(A=\frac{2^{201}-2}{3}\)
b) Ta có B = 3100- 399 + 398 - 397 + ... + 32 - 3 + 1
=> 3B = 3101 - 3100 + 399 - 398 + ... + 33 - 32 + 3
Khi đó 3B + B = (3101 - 3100 + 399 - 398 + ... + 33 - 32 + 3) + (3100- 399 + 398 - 397 + ... + 32 - 3 + 1)
=> 4B = 3101 + 1
=> B = \(\frac{3^{101}+1}{4}\)
a) \(A=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
=> \(2A=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
=> \(2A+A=\left(2^{101}-2^{100}+...-2^2\right)+\left(2^{100}-2^{99}+...-2\right)\)
<=> \(3A=2^{101}-2\)
=> \(A=\frac{2^{101}-2}{3}\)
b) \(B=3^{100}-3^{99}+3^{98}-3^{97}+...+3^2-3+1\)
=> \(3A=3^{101}-3^{100}+3^{99}-3^{98}+...+3^3-3^2+3\)
=> \(3A+A=\left(3^{101}-3^{100}+...+3\right)+\left(3^{100}-3^{99}+...+1\right)\)
<=> \(4A=3^{101}+1\)
=> \(A=\frac{3^{101}+1}{4}\)
\(A=2^{100}-\left(2^{99}+2^{98}+.....+2^2+2+1\right)\)
\(2A=2^{101}-\left(2^{100}+2^{99}+......+2^3+2^2+2\right)\)
\(A=2^{101}-2^{100}+1\)
@coolkid Cảm ơn max nhiều :)