Question

[Hùng Nguyễn Thế]

Cho S = 3⁰ + 3²  +  3⁴ + 3⁶ + ... + 3²⁰⁰²

a) Tính S

b) Chứng minh S \(⋮\)7

Answer 1

[Phùng Minh Quân]

\(a)\) \(S=3^0+3^2+3^4+3^6+...+3^{2002}\)

\(9S=3^2+3^4+3^6+3^8+...+3^{2004}\)

\(9S-S=\left(3^2+3^4+3^6+3^8+...+3^{2004}\right)-\left(3^0+3^2+3^4+3^6+...+3^{2002}\right)\)

\(8S=3^{2004}-3^0\)

\(8S=3^{2004}-1\)

\(S=\frac{3^{2004}-1}{8}\)

Vậy \(S=\frac{3^{2004}-1}{8}\)

Answer 2

[Ngo Tung Lam]

a) \(S=3^0+3^2+3^4+3^6+....+3^{2002}\)

\(\Rightarrow3^2.S=3^2+3^4+3^6+3^8+....+3^{2004}\)

\(\Rightarrow9S-S=\left(3^2+3^4+3^6+3^8+....+3^{2004}\right)-\left(3^0+3^2+3^4+3^6+....+3^{2002}\right)\)

\(\Rightarrow8S=3^{2004}-1\)

\(\Rightarrow S=\frac{3^{2004}-1}{8}\)

Vậy \(S=\frac{3^{2004}-1}{8}\)

b) Ta có :

\(S=3^0+3^2+3^4+3^6+....+3^{2002}\)

Tổng \(S\)có số số hạng là :

( 2002 - 0 ) : 2 + 1 = 1002 ( số hạng )

Ta có : \(1002⋮3\)nên khi ta nhóm 3 số liên tiếp lại thành 1 nhóm thì sẽ không có số nào thừa cả 

\(\Rightarrow S=\left(3^0+3^2+3^4\right)+\left(3^6+3^8+3^{10}\right)+....+\left(3^{1998}+3^{2000}+3^{2002}\right)\)

\(\Rightarrow S=3^0\left(1+3^2+3^4\right)+3^6\left(1+3^2+3^4\right)+....+3^{1998}\left(1+3^2+3^4\right)\)

\(\Rightarrow S=1.91+3^6.91+....+3^{1998}.91\)

\(\Rightarrow S=91.\left(1+3^6+....+3^{1998}\right)\)

Vì \(1+3^6+....+3^{1998}\inℤ\)nên \(91.\left(1+3^6+....+3^{1998}\right)\inℤ\)

Vì \(91⋮7\)nên \(91.\left(1+3^6+....+3^{1998}\right)⋮7\)

Vậy \(S=3^0+3^2+3^4+3^6+....+3^{2002}⋮7\left(ĐPCM\right)\)