Question

[Nguyễn Văn Đông]

Cho S = 3^2 + 3^4 + ... + 3^998 + 3^1000

a) tính S

b) Chứng minh rằng S chia cho 7 dư 6

Answer 1

[Akai Haruma]

Lời giải:

$S=3^2+3^4+3^6+...+3^{998}+3^{1000}$

$3^2S=3^4+3^6+3^8+...+3^{1000}+3^{1002}$

$\Rightarrow 3^2S-S=3^{1002}-3^2$
$\Rightarrow 8S=3^{1002}-9$

$\Rightarrow S=\frac{3^{1002}-9}{8}$

b.

$S=3^2+3^4+(3^6+3^8+3^{10})+(3^{12}+3^{14}+3^{16})+...+(3^{996}+3^{998}+3^{1000})$

$=90+3^6(1+3^2+3^4)+3^{12}(1+3^2+3^4)+...+3^{996}(1+3^2+3^4)$

$=90+(1+3^2+3^4)(3^6+3^{12}+...+3^{996})$

$=90+91(3^6+3^{12}+...+3^{996})$

$=6+ 12.7+7.13(3^6+3^{12}+...+3^{996})$ chia $7$ dư $6$