\(S=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
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\(S=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(S=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(S=\left(100-99\right)\left(99+100\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(S=1\cdot199+1\cdot195+1\cdot193+...+1\cdot3\)
\(S=199+195+193+...+3\)
\(S=\left(3+199\right)\cdot\left[\left(199-3\right):2+1\right]:2\)
\(S=202\cdot99:2\)
\(S=101\cdot99\)
\(S=9999\)
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.
A=-1++(-1)+..+-(1) có 50 số -1
=>A=-1x50=-50
B=(1-2-3+4)+(5-6-7+8)+...+(97-98-99+100)
B=0+0+0+..+0
B=0
C=2^100-(2^99+2^98+...+1)
C=2^100-(2^100-1)
C=1
Lời giải:
$S=[(-1)+(-2)+3+4]+[(-5)+(-6)+7+8]+....+[(-97)+(-98)+99+100]$
$=\underbrace{4+4+....+4}_{25}=4.25=100$
b) B = 2100 - 299 + 298 - 297 + ...+ 22 - 2
=> B x 2 = 2101 - 2100 + 299 - 298 + ...23 - 22
=> B x 2 + B = (2101 - 2100 + 299 - 298 + ...23 - 22 ) + (2100 - 299 + 298 - 297 + ...+ 22 - 2)
<=> B x 3 = 2101 - 2 = 2. ( 299 - 1)
=> B = \(\frac{2.\left(2^{99}-1\right)}{3}\)
Phần c) Làm tương tự Lấy C x 3 rồi + với C.
A = 2100 - 299 + 298 - 297 +...+ 22 - 2
=> 2A = 2101 - 2100+299 - 298+...+23-22
=> 2A+A= 2101 -2
=> \(A=\frac{2^{101}-2}{3}\)
phần B bn lm tương tự nha!
\(S=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(2S=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(2S+S=\left(2^{101}-2^{100}+2^{99}-...-2^2\right)-\left(2^{100}-2^{99}+2^{98}-...-2\right)\)
\(3S=2^{101}-2\)
\(S=\frac{2^{101}-2}{3}\)
\(S=2^{100}-2^{99}+2^{98}-2^{97}+...+2^2-2\)
\(\Rightarrow\) \(2S=2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)
\(\Rightarrow\) \(3S=2^{101}-2\)
\(\Rightarrow\) \(S=\frac{2^{101}-2}{3}\)