* Chứng minh a chia hết cho 4

     a = 2 + $2^{2}$ + $2^{3}$ + ... +$2^{100}$ 

2a = 2.( 2 + $2^{2}$ + $2^{3}$ + ... +$2^{100}$ )

2a = $2^{2}$ + $2^{3}$ + $2^{4}$ + ... + $2^{101}$

2a = $2^{2}$. ( 1+$2^{1}$ +$2^{2}$ + ... + $2^{99}$ )

2a - a = $2^{2}$. ( 1+$2^{1}$ +$2^{2}$ + ... + $2^{99}$ ) - ( 2 + $2^{2}$ + $2^{3}$ + ... +$2^{100}$ )

a = $2^{2}$ . ( 1 + $2^{100}$)

a = 4 . ( 1 + $2^{100}$)

Vì 4 chia hết cho 4 nên 4. ( 1 + $2^{100}$) chia hết cho 4 

→ a chia hết cho 4

 Vậy a chia hết cho 4 

* Chứng minh a chia hết cho 5

    a = 2 + $2^{2}$ + $2^{3}$ + ... +$2^{100}$ 

a= ( 2 + $2^{2}$ + $2^{3}$ + $2^{4}$ ) + ( $2^{5}$ + $2^{6}$ + $2^{7}$ + $2^{8}$ ) + ( $2^{9}$ + $2^{10}$ + $2^{11}$ + $2^{12}$ ) + ... + ( $2^{97}$ + $2^{98}$ + $2^{99}$ + $2^{100}$ )

a = 2. (1+ 2 + 2^2 + 2^3) + $2^{5}$.(1+ 2 + 2^2 + 2^3) + ... + $2^{97}$ .(1+ 2 + 2^2 + 2^3)

a = (2 + $2^{5}$ + $2^{9}$ +...+ $2^{97}$). 15

Vì 15= 3.5 mà 5 chia hết cho 5 nên (2 + $2^{5}$ + $2^{9}$ +...+ $2^{97}$). 15 chia hết cho 5

→ a chia hết cho 5 

 Vậy a chia hết cho 5

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