\(A=3^{100}-3^{99}+3^{98}-...-3+1\\ \Rightarrow\dfrac{1}{3}A=3^{99}-3^{98}+3^{97}-...-1+\dfrac{1}{3}\\ \Rightarrow\dfrac{4}{3}A=3^{100}+\dfrac{1}{3}\\ \Rightarrow A=\dfrac{3^{101}}{4}+\dfrac{1}{4}\)

=1+(2-3-4+5)+(6-7-8+9)+...+(398-399-400+401)+402

=(2+5-3-4)+(6+9-7-8)+...+(398+401-399-400)+1+402

=0+0+...+0+403

=403

h mình nha

gfdgdfyrdgd

ta có số các số hạng là 398-1+1=398 số hạng

a) A=(1-2)+(3-4)+(5-6)+.......+(397-398)

A=(-1)+(-1)+.....+(-1)

có 398/2=199 cặp

vậy A=(-1)*199=-199

\(A=1-2+3-4+5-6+...+397-398\)

\(A=\left(1-2\right)+\left(3-4\right)+\left(5-6\right)+...+\left(397-398\right)\)

\(A=\left(-1\right)+\left(-1\right)+\left(-1\right)+...\left(-1\right)\)

\(A=\left(-1\right)\cdot199\)

\(A=-199\)

\(B=1+3-5-7+9+11-...+393-395-397-399\)( chỗ này mình cố ý viết thêm để dễ nhìn ) 

\(B=1+\left(3-5-7+9\right)-\left(11-13-15+17\right)-...-\left(387-389-391+393\right)-\left(395-397-399\right)\)

\(B=1+0-0-...-0-\left(-401\right)\)

\(B=1-\left(-401\right)\)

\(B=402\)

a) Ta có: \(\dfrac{25^{28}+25^{24}+25^{20}+...+25^4+1}{25^{30}+25^{28}+...+25^2+1}\)

\(=\dfrac{25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+...+\left(25^4+1\right)}{25^{28}\left(25^2+1\right)+25^{24}\left(25^2+1\right)+...+\left(25^2+1\right)}\)

\(=\dfrac{\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}{\left(25^2+1\right)\left(25^{28}+25^{24}+...+1\right)}\)

\(=\dfrac{\left(25^4+1\right)\cdot\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}{\left(25^2+1\right)\left[25^{24}\left(25^4+1\right)+25^{16}\left(25^4+1\right)+25^8\left(25^4+1\right)+\left(25^4+1\right)\right]}\)

\(=\dfrac{\left(25^4+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^4+1\right)\left(25^{24}+25^{16}+25^8+1\right)}\)

\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left[25^{16}\left(25^8+1\right)+\left(25^8+1\right)\right]}\)

\(=\dfrac{\left(25^8+1\right)\left(25^{16}+1\right)}{\left(25^2+1\right)\left(25^8+1\right)\left(25^{16}+1\right)}\)

\(=\dfrac{1}{25^2+1}=\dfrac{1}{626}\)

A = 1 + 3 + 5 + 7 + .......... + 397 + 399

A có số số hạng là:

( 399 - 1 ) : 2 + 1 = 200 ( số hạng )

A có kết quả là:

( 399 + 1 ) x 200 : 2 = 40000

B = 2 + 4 + 6 + 8 + .......... + 396 + 398

B có số số hạng là:

( 398 - 2 ) : 2 + 1 = 199 ( số hạng )

B có kết quả là:

( 398 + 2 ) x 199 : 2 = 39800

Vì 40000 > 39800 nên A > B

tks : D tớ cx đang định làm  như thế : D
 

\(A=1+3+5+....+399=\frac{\left[\left(399-1\right):2+1\right]\left(399+1\right)}{2}=39000\)

\(B=2+4+.....+398=\frac{\left[\left(398-2\right):2+1\right]\left(398+2\right)}{2}=39850\)

Vậy ..

Tham khảo

Ta có: 3A = 3.(1+3+32+33+...+399+3100)(1+3+32+33+...+399+3100)

3A = 3+32+33+...+3100+31013+32+33+...+3100+3101

Suy ra: 3A – A = (3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)(3+32+33+...+3100+3101)−(1+3+32+33+...+399+3100)

2A = 3101−13101−1

⇒⇒ A = 3101−123101−12

Vậy A = 3101−12

A=3 mũ 101-1 phân số2

\(A=1-3+3^2-3^3+3^4-...-3^{98}-3^{99}+3^{100}\\ 3A=3-3^2+3^3-3^4-...-3^{98}+3^{99}-3^{100}+3^{101}\\ 3A-A=3^{101}-1\\ \Rightarrow A=\dfrac{3^{101}-1}{2}\)

       A =  1 - 3 +  3² -   3³ + 3⁴ - ... + 3⁹⁸ - 3⁹⁹ + 3¹⁰⁰

      3A =  3 - 3² + 3³ - 3⁴+ 3⁵  - ... + 3⁹⁹ - 3¹⁰⁰ + 3¹⁰¹

3A + A = 3 - 3²+ 3³-3⁴+3⁵ -...+3⁹⁹ - 3¹⁰⁰ + 3¹⁰¹ + 1 - 3 +...-3⁹⁹+3¹⁰⁰

4A   =    3¹⁰¹ + 1

  A    = \(\dfrac{3^{101}+1}{4}\)