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:

$X=1.2+2.3+3.4+....+100.101$

$3X=1.2(3-0)+2.3(4-1)+3.4(5-2)+....+100.101(102-99)$

$=3X=(1.2.3+2.3.4+3.4.5+....+100.101.102)-(0.1.2+1.2.3+...+99.100.101)$

$=100.101.102$

$\Rightarrow X=\frac{100.101.102}{3}$

$Y=1+2+3+...+100=100(100+1):2=5050$ 

$A=X-Y=\frac{100.101.102}{3}-5050=338350$

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

Bài 1 :

\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)

Ta thấy :

\(3=2^2-1\)

\(15=4^2-1\)

\(35=6^2-1\)

.....

\(9999=100^2-1\)

\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)

\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)

\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)

nhanh len nhé mik đang cần gấp ai lam trước mik tích cho

Bài 6 :

\(C=1^2+2^2+...+100^2=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}=\dfrac{100.101.201}{6}=338350\)

Bài 9 :

\(S=1^2+2^2+3^2+...+99^2=\dfrac{99.\left(99+1\right)\left(2.99+1\right)}{6}=\dfrac{99.100.199}{6}=328350\)

1+2+3+4+...+100

100+99+98+97+...+1

=101+101+101+101+...+101):2

=10100:2

=5050

bang5050 do k cho minh nho