1,tính tổng S=1+2^2+2^4+2^6+…….+2^98+2^100 2, tính tổng S=6^2+6^4+6^6+………+6^98++^100 3, tính tổng S=1+3^2+3^4+3^6+….+3^100+3^102

0 bình luận về “1,tính tổng S=1+2^2+2^4+2^6+…….+2^98+2^100 2, tính tổng S=6^2+6^4+6^6+………+6^98++^100 3, tính tổng S=1+3^2+3^4+3^6+….+3^100+3^102”

1. $1$.

   $S =  1 + 2^2 + 2^4 + 2^6 + …. + 2^{98} + 2^{100}$

   $⇔ 2^2S = 2^2 + 2^4 + 2^6 + 2^8 + …. + 2^{100} + 2^{102}$

   $⇔ 4S – S = (2^2 + 2^4 + 2^6 + 2^8 + …. + 2^{100} + 2^{102})-( 1 + 2^2 + 2^4 + 2^6 + …. + 2^{98} + 2^{100})$

   $⇔ 3S = 2^{102} -1$

   $⇔ S = \dfrac{2^{102}-1}{3}$

   $2$.

   $S = 6^2 + 6^4 + 6^6 + …. + 6^{98} + 6^{100}$

   $⇔ 6^2S = 6^4 + 6^6 + 6^8 + …. + 6^{100} +  6^{102}$

   $⇔ 36S – S = (6^4 + 6^6 + 6^8 + …. + 6^{100} +  6^{102})-(6^2 + 6^4 + 6^6 + …. + 6^{98} + 6^{100})$

   $⇔ 35S = 6^{102} – 6^2$

   $⇔ 35S = \dfrac{6^{102}-6^2}{35}$

   $3$. 

   $S = 1 + 3^2 + 3^4 + 3^6 + …. + 3^{100} + 3^{102}$

   $⇔ 3^2S = 3^2 + 3^4 + 3^6 + 3^8 + …. + 3^{102} + 3^{104}$

   $⇔ 9S – S = (3^2 + 3^4 + 3^6 + 3^8 + …. + 3^{102} + 3^{104})-(1 + 3^2 + 3^4 + 3^6 + …. + 3^{100} + 3^{102})$

   $⇔ 8S = 3^{104} – 1$

   $⇔ S = \dfrac{3^{104}-1}{8}$

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2. 1 : 

   $S = 1 + 2^{2} +…+ 2^{100}$ 

   $4S = 2^{2} + 2^{4} +…+ 2^{102}$ 

   $4S – S = (2^{2} + 2^{4} +…+ 2^{102}) – (1 + 2^{2} +…+ 2^{100})$

   $3S = 2^{102} – 1$

   $S = \frac{2^{102} – 1}{3}$ 

   2 : 

   $S = 6^{4} + 6^{6} +…+ 6^{102}$

   $36S = 6^{4} + 6^{6} +…+ 6^{102}$

   $36S – S = (6^{4} + 6^{6} +…+ 6^{102}) – (6^{2} + 6^{4} +…+ 6^{100})$

   $35S = 6^{102} – 6^{2}$

   $S = \frac{6^{102} – 6^{2}}{35}$

   3 :

   $S = 1 + 3^{2} +….+ 3^{102}$

   $9S = 3^{2} + 3^{4} +….+ 3^{104}$

   $9S – S = (3^{2} + 3^{4} +….+ 3^{104}) – (1 + 3^{2} +….+ 3^{102})$

   $9S = 3^{104} – 1$

   $S = \frac{3^{104} – 1}{9}$

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