A = 5 + 5 ^ 2 + 5 ^ 3 + ... + 5 ^ 50

5 A = 5 ^ 2 + 5 ^ 3 + 5 ^ 4 + ... + 5 ^ 51

5 A - A = ( 5 ^ 2 + 5 ^ 3 + 5 ^ 4 + ... + 5 ^ 51 )

            -  ( 5 + 5 ^ 2 + 5 ^ 3 + ... + 5 ^ 50 )

4 A      = 5 ^ 51 - 5

A         = \(\frac{5^{51}-5}{4}\)

A=5^1+5^21+5^3+...+5^50

5^1A=5(5^1+5^2+5^3+..+5^50)

5A=5^2+5^3+..+5^50+5^51

5A-A=(5^2+5^3+..+5^50+5^51)-(5^1+5^2+5^3+..+5^50)

4A=5^51-5^1

A=(5^51-5^1):4

=(5^51-5):25

\(A=2^0+2^1+2^2\)\(+2^3+...+\)\(2^{50}\)

\(2A=2+2^2+2^3+...+2^{51}\)

\(2A-A=A=2^{51}-2^0\)

\(B=5+5^2+5^3+...+5^{99}+5^{100}\)

\(5B=5^2+5^3+5^4+...+5^{100}+5^{101}\)

\(5B-B=4B=5^{101}-5\)

\(B=\frac{5^{101}-5}{4}\)

\(C=3-3^2+3^3-3^4+...+\)\(3^{2007}-3^{2008}+3^{2009}-3^{2010}\)

\(3C=3^2-3^3+3^4-3^5+...-3^{2008}+3^{2009}-3^{2010}+3^{2011}\)

\(3C+C=4C=3^{2011}+3\)

\(C=\frac{3^{2011}+3}{4}\)

\(S_{100}=5+5\times9+5\times9^2+5\times9^3+...+5\times9^{99}\)

\(S_{100}=5\times\left(1+9+9^2+9^3+...+9^{99}\right)\)

\(9S_{100}=5\times\left(9+9^2+9^3+...+9^{99}+9^{100}\right)\)

\(9S_{100}-S_{100}=8S_{100}=5\times\left(9^{100}-1\right)\)

\(S_{100}=\frac{5\times\left(9^{100}-1\right)}{8}\)

$A=2^0+2^1+2^2$$+2^3+...+$$2^{50}$

$2A=2+2^2+2^3+...+2^{51}$

$2A-A=A=2^{51}-2^0$

$B=5+5^2+5^3+...+5^{99}+5^{100}$

$5B=5^2+5^3+5^4+...+5^{100}+5^{101}$

$5B-B=4B=5^{101}-5$

$B=\genfrac{}{}{0ex}{}{5^{101}-5}{4}$

$C=3-3^2+3^3-3^4+...+$$3^{2007}-3^{2008}+3^{2009}-3^{2010}$

$3C=3^2-3^3+3^4-3^5+...-3^{2008}+3^{2009}-3^{2010}+3^{2011}$

$3C+C=4C=3^{2011}+3$

$C=\genfrac{}{}{0ex}{}{3^{2011}+3}{4}$

$S^{100}=5+5\times 9+5\times 9^2+5\times 9^3+...+5\times 9^{99}$

$S^{100}=5\times \left(1+9+9^2+9^3+...+9^{99}\right)$

$9S^{100}=5\times \left(9+9^2+9^3+...+9^{99}+9^{100}\right)$

$9S^{100}-S^{100}=8S^{100}=5\times \left(9^{100}-1\right)$

$S^{100}=\genfrac{}{}{0ex}{}{5\times \left(9^{100}-1\right)}{8}$

\(A=1+5+5^2+..+5^{49}+5^{50}\)

\(5A=5+5^2+5^3+...+5^{50}+5^{51}\)

\(5A-A=\left(5+5^2+5^3+...+5^{51}\right)-\left(1+5+5^2+...+5^{50}\right)\)

\(4A=\left(5-5\right)+\left(5^2-5^2\right)+...+\left(5^{50}+5^{50}\right)+5^{51}-1\)

\(4A=0+0+...+0+5^{51}-1\)

\(4A=5^{51}-1\)

\(A=\frac{5^{51}-1}{4}\)

Ai nhanh mik k thề đó

 Thu gọn : 5 + 5^2 + 5^3 + ... + 5^49 + 5^50

đặt tên biểu thức trên là A

ta có :

\(A=5+5^2+5^3+...+5^{49}+5^{50}\)

\(5A=5.\left(5+5^2+5^3+...+5^{49}+5^{50}\right)\)

\(5A=5^2+5^3+5^4+...+5^{50}+5^{51}\)

\(5A-A=\left(5^2+5^3+5^4+...+5^{50}+5^{51}\right)-\left(5+5^2+5^3+...+5^{49}+5^{50}\right)\)

\(4A=5^{51}-5\)

\(A=\left(5^{51}-5\right):4\)