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S = 1.2.3.4 + 2.3.4.5 + 3.4.5.6+...97.98.99.100
5S = (1.2.3.4+2.3.4.5+3.4.5.6+ ... + 97.98.99.100).5
5S = 1.2.3.4.(5-0) + 2.3.4.5.(6-1)+ 3.4.5.6(7-2)+......+ 97.98.99.100.(101-96)
5S = (1.2.3.4.5 + 2.3.4.5.6 + 3.4.5.6.7 + ....+ 97.98.99.100.101) - (0.1.2.3.4 + 1.2.3.4.5 + 2.3.4.5.6+.....+96.97.98.99.100)
5S = 97.98.99.100.101
S= 97.98.99.100.101/5
S=1901009880
S=1*2*3*4+2*3*4*5+....+97*98*99*100
5S=1.2.3.4.5+2.3.4.5.5+...+97.98.99.100.5
5S=1.2.3.4.(5-0)+2.3.4.5.(6-1)+...+97.98.99.100.(101-96)
5S=1.2.3.4.5-0.1.2.3.4+2.3.4.5.6-1.2.3.4.5+...+97.98.99.100.101-96.97.98.99.100
5S=(1.2.3.4.5+2.3.4.5.6+...+97.98.99.100.101)-(0.1.2.3.4+1.2.3.4.5+...+96.97.98.99.100)
5S=97.98.99.100.101
S=9505049400:5=1901009880.
Mình không chắc là có đúng không nữa các bạn xem hộ mình với nha!
= (100^2 - 99^2) + (98^2 - 97^2) + ... + (4^2 - 3^2) + (2^2 - 1^2) =
= (100+99)(100-99) + (98+97)(98-97) + ... + (4+3)(4-3) + (2+1)(2-1) =
= (100+99).1 + (98+97).1 + ... + (4+3).1 + (2+1).1 =
= 100 + 99 + 98 + 97 + ... + 4 + 3 + 2 + 1 =
= (100+1) + (99+2) + (98+3) + ... + (51+50) = 101.50 = 5050
(50 cặp dấu ngoặc, tổng trong mỗi cặp dấu ngoặc là 101)
Bài 1: Tính nhanh
a) Ta có: \(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+97+...+2+1\)
\(=\left(100+1\right)+\left(99+2\right)+\left(98+3\right)+\left(97+4\right)+...+\left(50+51\right)\)
\(=101\cdot50=5050\)
b) Ta có: \(B=\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow4\cdot B=24\cdot\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow4\cdot B=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow4\cdot B=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow4\cdot B=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow4\cdot B=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow4\cdot B=5^{32}-1\)
hay \(B=\frac{5^{32}-1}{4}\)
\(\left(100^2+98^2+...+2^2\right)-\left(99^2+97^2+...+1^2\right)\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+....+\left(2^2-1^2\right)\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+97+....+2+1=5050\)
\(\left(100^2+98^2+...+2^2\right)-\left(99^2+97^2+...+1^2\right)\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+....+\left(2^2-1^2\right)\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+97+....+2+1=5050\)
a) Ta có : $1.3+2.4+3.5+...+99.101+100.102$
$=(2-1)(2+1)+(3-1)(3+1)+(4-1)(4+1)+...+(100-1)(100+1)+(101-1)(101+1)$
$=2^2-1+3^2-1+4^2-1+...+100^2-1+101^2-1$
$=(2^2+3^2+4^2+...+100^2+101^2)-100$
b) $1.100+2.99+3.98+...+99.2+100.1$
$=1.100+2.(100-1)+3.(100-2)+...+99.(100-98)+100.(100-99)$
$=100(1+2+3+...+99+100)-(1.2+2.3+...+99.100)$
$=100.\dfrac{101.100}{2}-\dfrac{99.100.101}{3}=171700$
\(1+2+....+2^{99}=2\left(1+2+....+2^{99}\right)-1-2-....-2^{99}=2^{100}-1\)
\(\Rightarrow2^{100}-\left(1+2+....+2^{99}\right)=2^{100}-\left(2^{100}-1\right)=1\)
Đặt biểu thức đã cho là A
\(\Rightarrow A=2^{100}-\left(2^{99}+2^{98}+2^{97}+......+2^2+2+1\right)\)
Đặt \(B=2^{99}+2^{98}+2^{97}+.......+2^2+2+1\)
\(\Rightarrow2B=2^{100}+2^{99}+2^{98}+.........+2^3+2^2+2\)
\(\Rightarrow2B-B=B=2^{100}-1\)
\(\Rightarrow A=2^{100}-B=2^{100}-\left(2^{100}-1\right)=2^{100}-2^{100}+1=1\)
\(1^2-2^2+3^2-4^2+...+97^2-98^2+99^2-100^2=\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+...+\left(97-98\right)\left(97+98\right)+\left(99-100\right)\left(99+100\right)\)\(=-\left(1+2+3+4+...+97+98+99+100\right)\)
\(=-\left(\frac{101\times100}{2}\right)=-5050\)
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