Đặt A = 1 + $2^{2}$ + $2^{4}$ + $2^{6}$ + ... + $2^{98}$ + 2$2^{100}$
⇒ 4A = 2² + $2^{4}$ + $2^{6}$ + ... + $2^{102}$
⇒ 4A - A = ( 2² + $2^{4}$ + $2^{6}$ + .... + $2^{102}$ ) - ( 1 + 2² + $2^{4}$ + $2^{6}$ + ... +$2^{100}$ )
⇒ 3A = $2^{102}$ - 1
⇒ A = $\frac{2^{102}-1}{3}$
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