Lời giải:

$A=\underbrace{(100+98+96+....+2)}_{M}-\underbrace{(99+97+....+1)}_{N}$

Tổng số hạng của $M$: $(100-2):2+1=50$

$M=(100+2).50:2=2550$

Tổng số hạng của $N$: $(99-1):2+1=50$

$N=(99+1).50:2=2500$

$A=M-N=2550-2500=50$

Sửa đề: A=100+98+96+...+2-99-97-...-1

=100-99+98-97+...+2-1

=1+1+...+1

=50

Ta có:

\(B=2\cdot\left(1\cdot99+2\cdot98+...+50\cdot50\right)-50\cdot50\)

\(=2\cdot\left(1\cdot99+2\cdot\left(99-1\right)+...+50\cdot\left(99-49\right)\right)-50\cdot50\)-

\(=2\cdot\left(1\cdot99+2\cdot99-1\cdot2+...+50\cdot99-49\cdot50\right)-50\cdot50\)

\(=2\cdot\left(\left(1\cdot99+2\cdot99+...+50\cdot99\right)-\left(1\cdot2+2\cdot3+...+49\cdot50\right)\right)-50\cdot50\)

\(=2\cdot\left(\frac{99\cdot50\cdot51}{2}-\frac{49\cdot50\cdot51}{3}\right)-50\cdot50\)

\(=2\cdot84575-2500\)

\(=166650\)

Vậy B=166650

A=1.99+2.98+3.97+...+97.3+98.2+99.1
A=1.99+2.(99−1)+3.(99−2)+...+98.(99−97)+99.(99−98)
A=1.99+2.99−1.2+3.99−2.3+98.99−97.98+99.99−98.99
=(1.99+2.99+3.99+...+98.99+99.99)−(1.2+2.3+3.4+...+97.98+98.99)
=99.(1+2+3+...+98+99)−(1.2+2.3+3.4+...+97.98+98.99)
=99.4950−(1.2+2.3+3.4+97.98+98.99)
Mà 1.2+2.3+3.4+...97.98+98.99
=​ 1/3 ​.[1.2+2.3.(4−1)+3.4.(5−2)+98.99.(100−97)]
=1/3​​.98.99.100

=323400
⇒A=99.4950−323400=166650

166650 nha bạn 

1-2-3+4+5-....+96+97-98-99+100

=(1-2-3+4)+....+(97-98-99+100)

=0+.........+0

=0

1 - 2 - 3 + 4 + 5 - 6 - 7 + 8 + ... + 96 + 97 - 98 - 99 + 100

Ta thấy tổng trên có 100 số hạng. Ta chia tổng thành tường nhóm, mỗi nhóm có 4 số hạng như sau:

1 - 2 - 3 + 4 + 5 - 6 - 7 - 8 + ... + 96 + 97 - 98 - 99 + 100

= ( 1 - 2 - 3 + 4 ) + ( 5 - 6 - 7 + 8 ) + ... + ( 97 - 98 - 99 + 100 )

= 0 + 0 + ... + 0

= 0

\(A=2^{100}-2^{99}+2^{98}-2^{97}+2-1\\ 2A=2^{101}-2^{100}+2^{99}+...+2^2-2\\ 2A+A=\left(2^{101}-2^{100}+2^{99}+...+2^2-2\right)+\left(2^{100}-2^{99}+2^{98}-2^{97}+2-1\right)\\ 3A=2^{101}-1\\ A=\dfrac{2^{101}-1}{3}\)

\(B=3+3^2+3^3+...+3^{2017}\\ 3B=3^2+3^3+3^4+...+3^{2018}\\ 3B-B=\left(3^2+3^3+3^4+...+3^{2018}\right)-\left(3+3^2+3^3+...+3^{2017}\right)\\ 2B=3^{2018}-3\\ B=\dfrac{3^{2018}-3}{2}\)

\(C=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{1000}}\\ 2C=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{999}}\\ 2C-C=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{999}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{1000}}\right)\\ C=1-\dfrac{1}{2^{1000}}\)

bằng 1

\(A=2^{100}-2^{99}+2^{98}-2^{97}+....+2^2-2\)

\(2A=2^{101}-2^{100}+2^{99}-2^{98}+....+2^3-2^2\)

\(2A+A=2^{101}-2\)

\(A=\frac{2^{101}-2}{3}\)

b) tương tự

\(B=\frac{3^{101}+1}{4}\)