tinh tong A = 1/3+1/362+..+1/3^100
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Lời giải:
$A=1.1+2.2+3.3+...+100.100$
$=1(2-1)+2(3-1)+3(4-1)+...+100(101-1)$
$=1.2+2.3+3.4+....+100.101-(1+2+3+...+100)$
Có:
$X=1.2+2.3+3.4+....+100.101$
$3X=1.2(3-0)+2.3(4-1)+3.4(5-2)+....+100.101(102-99)$
$=3X=(1.2.3+2.3.4+3.4.5+....+100.101.102)-(0.1.2+1.2.3+...+99.100.101)$
$=100.101.102$
$\Rightarrow X=\frac{100.101.102}{3}$
$Y=1+2+3+...+100=100(100+1):2=5050$
$A=X-Y=\frac{100.101.102}{3}-5050=338350$
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bai1 tu -1738 den -16 co so cac so nguyen am la : (1738-16)/2+1=862
C1
B=1+3+5+..+99=50.(99+1)/2=50.50=2500
C=-(2+4+6+...+100)=50.(100+2)/2=50*51=-2550
A=B+C=2500-2550=-50
C2
A=1+(3-2)+(5-4)+...+(99-98)+100
=1+1+1+...+1-100
=1+49-100=-50
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Bài 1 :
\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)
Ta thấy :
\(3=2^2-1\)
\(15=4^2-1\)
\(35=6^2-1\)
.....
\(9999=100^2-1\)
\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)
\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)
\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)
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1+2+3+4+...+100
100+99+98+97+...+1
=101+101+101+101+...+101):2
=10100:2
=5050