Question

Tính tổng sau :

Cho A = 2 mũ 6 + 2 mũ 8 +...........+ 2 mũ 98 + 2 mũ 100

a) Tính A

b) Chứng minh rằng A chia hết cho 5

Answer 1

Lời giải:

a)

\(A=2^6+2^8+...+2^{98}+2^{100}\)

\(\Rightarrow 2^2A=2^8+2^{10}+...+2^{100}+2^{102}\)

Trừ theo vế:

\(4A-A=(2^8+2^{10}+..+2^{100}+2^{102})-(2^6+2^8+...+2^{98}+2^{100})\)

\(3A=2^{102}-2^6\)

\(\Rightarrow A=\frac{2^{102}-2^6}{3}\)

b)

\(A=2^6+2^8+2^{10}+...+2^{98}+2^{100}\)

\(A=(2^6+2^8)+(2^{10}+2^{12})+...+(2^{98}+2^{100})\)

\(=2^6(1+2^2)+2^{10}(1+2^2)+...+2^{98}(1+2^2)\)

\(=5.2^6+5.2^{10}+...+5.2^{98}=5(2^6+2^{10}+...+2^{98})\vdots 5\)

Ta có đpcm.