a) Ta có : C = 3² + 3⁴ + 3⁶ + ... + 3²⁰² + 3²⁰⁴

=> 3²C = 9C = 3⁴ + 3⁶ + 3⁸ + .... + 3²⁰⁴ + 3²⁰⁶

Lấy 9C trừ C theo vế ta có 

9C - C = (3⁴ + 3⁶ + 3⁸ + .... + 3²⁰⁴ + 3²⁰⁶) - ( 3² + 3⁴ + 3⁶ + ... + 3²⁰² + 3²⁰⁴)

=> 8C = 3²⁰⁶ - 3²

=> C = \(\frac{3^{206}-3^2}{8}\)

d) Ta có D = \(\frac{1}{3^2}-\frac{1}{3^4}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\)

=> 3²D = 9D = \(1-\frac{1}{3^2}+...+\frac{1}{3^{200}}-\frac{1}{3^{202}}\)

Lấy 9D cộng D theo vế ta có :

9D + D = \(\left(1-\frac{1}{3^2}+...+\frac{1}{3^{200}}-\frac{1}{3^{202}}\right)+\left(\frac{1}{3^2}-\frac{1}{3^4}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\right)\)

=> 10D = \(1-\frac{1}{3^{204}}\)

=> D = \(\frac{1}{10}-\frac{1}{3^{204}.10}\)

thiếu đề rồi cố nội, sửa lại đê

Số số hạng là:

(204-2):2+1=102(số hạng)

Tổng:

(2-4)+(6-8)+...+(202-204)

=(-2)+(-2)+...+(-2)

=(-2)x102

=-204

\(A=\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\)

\(9A=\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^4}-\frac{1}{3^6}+...+\frac{1}{3^{200}}-\frac{1}{3^{202}}\)

\(9A+A=\left(\frac{1}{3}-\frac{1}{3^{^2}}+...+\frac{1}{3^{200}}-\frac{1}{3^{202}}\right)+\left(\frac{1}{3^2}-\frac{1}{3^4}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\right)\)

\(10A=\frac{1}{3}-\frac{1}{3^{204}}\)

A = (1/3 - 1/3²⁰⁴) : 10

Vậy A = (1/3 - 1/3²⁰⁴) : 10.

A= \(\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\left(1\right)\\ \)

   \(\frac{1}{3^2}A=\frac{1}{3^2}\left(\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\right)\)

    \(\frac{1}{3^2}A=\frac{1}{3^4}-\frac{1}{3^6}+\frac{1}{3^8}-\frac{1}{3^{10}}+...+\frac{1}{3^{204}}-\frac{1}{3^{206}}\left(2\right)\)

Từ (1) và (2) vế theo vế ta có :\(A-\frac{1}{3^2}A=\frac{8}{9}A=\frac{1}{3^2}-\frac{1}{3^{206}}\)

        \(\Rightarrow A=\left(\frac{1}{3^2}-\frac{1}{3^{206}}\right):\frac{8}{9}\)

\(A=\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}+\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\)

\(\frac{1}{3^2}A=\frac{1}{3^2}\left(\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{202}}-\frac{1}{3^{204}}\right)\)

\(\frac{1}{3^2}A=\frac{1}{3^4}-\frac{1}{3^6}+\frac{1}{3^8}-\frac{1}{3^{10}}+...+\frac{1}{3^{204}}-\frac{1}{3^{206}}\)

Do đó , \(A-\frac{1}{3^2}A=\frac{8}{9}A=\frac{1}{3^2}-\frac{1}{3^{206}}\)

\(\Rightarrow A=\left(\frac{1}{3^2}-\frac{1}{3^{205}}\right)\div\frac{8}{9}\)

vì 204 chia hết 4 ta ghép 4 số liên tiếp lại một cặp sau đó  được bao nhiêu mổi cặp rồi nhân lên

Bằng -204

Tính các tổng sau:

1, S=1-2+3_4+..+25-26

S =-1+3-5+7-...-53+55                       ( có 28 số hạng )

   = (-1+3)+(-5+7)+...+(-53+55)         ( có 28:2=14 nhóm )

   = 2+2+...+2

    = 2 . 14

     = 28

Á nhầm rùi xl bn nha

3 

S =-1+3-5+7-...-53+55     ( có 28 số hạng )

   = (-1+3)+(-5+7)+...+(-53+55)      ( có 28:2=14 nhóm )

   = 2+2+...+2

    = 2 . 14

     = 28

=1+(2-3-4+5)+3-4-5+6)+...+(200-201-202+203)+204

=1+0+0+...+0+204

=1+204

=205

tinh 

1+2 -3 -4 +5+3-4-5+6 +4-5-6 +7 + ....+200-201 -202 + 203 + 204 =1+0+...+0+204=1+204=205.

bn k cho mik nha. ^-^ thanks bn trc.

ban oi minh chua hoc so am 

A=(5-1)+(13-9)+...+(85-81)=4+4+...+4=4*11=44

B=(-4+2)+(-8+6)+...+(-204+202)=(-2)+(-2)+...+(-2)=(-2)*51=-102

Tính ._.

\(A=3^2+3^4+3^6+...+3^{204}\)

\(9A=3^4+3^6+3^8+...+3^{206}\)

\(9A-A=\left(3^4+3^6+3^8+...+3^{206}\right)-\left(3^2+3^4+3^6+...+3^{204}\right)\)

\(8A=3^{206}-3^2\)

\(\Rightarrow A=\frac{3^{206}-3^2}{8}\)

Vậy \(A=\frac{3^{206}-3^2}{8}\).

2:

a: =4+3/8+5+2/3

=9+3/8+2/3

=216/24+9/24+16/24

=216/24+25/24

=241/24

b; =2+3/8+1+1/4+3+6/7

=6+3/8+1/4+6/7

=6+5/8+6/7

=419/56

c: \(=2+\dfrac{3}{8}-1-\dfrac{1}{4}+5+\dfrac{1}{3}\)

=6+3/8-1/4+1/3

=6+1/8+1/3

=6+11/24

=155/24

d: \(=3+\dfrac{5}{6}+6\cdot\dfrac{13}{6}\)

=3+13+5/6

=16+5/6

=101/6

e: =3+1/2+4+5/7-5-5/14

=3+4-5+1/2+5/7-5/14

=2+7/14+10/14-5/14

=2+12/14

=2+6/7=20/7

f: =9/2+1/2:11/2

=9/2+1/11

=99/22+2/22=101/22