1) $\text{M = 1 + 3 + 5 + ... + 97 + 99}$

               Giải

$\text{M có số số hạng là:}$

$\text{(99 - 1) : 2 + 1 = 50 (số số hạng)}$

$\text{Tổng của M là:}$

$\text{(99 + 1) . 50 : 2 = 2500}$

                      $\text{Đ/S: 2500}$

2) $\text{B = 1 + 2 + $2^{2}$ + $2^{3}$ + ... + $2^{2007}$}$ 

  $\text{2B = 2 + $2^{2}$ + $2^{3}$ + ... + $2^{2007}$ + $2^{2008}$}$ 

$\text{2B - B = 2+$2^{2}$ + $2^{3}$+ ... + $2^{2007}$ + $2^{2008}$ - (1 + 2 + $2^{2}$ + $2^{3}$}$ $\text{+ ... + $2^{2007}$)}$

  $\text{B= 2 + $2^{2}$ + $2^{3}$ + ... + $2^{2007}$ + $2^{2008}$ - 1 - 2 - $2^{2}$ - $2^{3}$ - ... -}$ $\text{$2^{2007}$}$

      $\text{= $2^{2008}$ - 1 => (đpcm)}$

1) M=1+3+5+...+97+99

M.2=(99+1)+(97+3)+...+(1+99)

M.2=100+100+...+100

M.2=100.50

M.2=5000

M=5000:2

M=2500

2) B=1+$2^{2}$ +$2^{3}$+...+$2^{2007}$

B.2=1+$2^{2}$ +$2^{3}$+...+$2^{2007}$

B.2=2+$2^{2}$ +$2^{3}$+...+$2^{2008}$

(B.2)-B=(2+$2^{2}$ +$2^{3}$+...+$2^{2008}$)-(1+$2^{2}$ +$2^{3}$+...+$2^{2007}$)

⇒$2^{2008}$)-1