Question

Cho S=3^0+3^2+3^4+.......+3^2000+3^2002

a) Tính S

b) Chứng minh S chia hết cho 7

Trình bày luôn nha nếu thế mình tick cho

OK

Answer 1

a)S=3^0+3^2+3^4+...+3^2000+3^2002

=>3^2S=3^2(3^0+3^2+3^4+...+3^2000+3^2002)

=>9S=3^2+3^4+3^6+...+3^2002+3^2004

=>9S-S=(3^2+3^4+3^6+...+3^2004)-(3^0+3^2+3^4+...+3^2000+3^2002)

=>8S=3^2004-3^0=3^2004-1

=>S=(3^2004-1)/8

b) S=3^0+3^2+3^4+...+3^2000+3^2004

=>S=(3^0+3^2+3^4)+(3^6+3^8+3^10)+...+(3^1998+3^2000+3^2002)

=>S=(1+3^2+3^4)+3^6(1+3^2+3^4)+...+3^1998(1+3^2+3^4)

=>S=91+3^6.91+...+3^1998.91

=>S=91(1+3^6+...+3^1998)

=>S=7.13.(1+3^6+...+3^1998

=>S chia hết cho 7 

Answer 2

b)Ta có:S=(3⁰+3²+3⁴)+...(3¹⁹⁹⁶+3¹⁹⁹⁸+3²⁰⁰⁰+3²⁰⁰²)

           S=91+...+3¹⁹⁹⁶.(1+3²+3⁴)

          S=91+...+3¹⁹⁹⁶.91

          S=91.(1+...+3¹⁹⁹⁶)

Vì 91chia hết cho 7 nên S chia hết cho 7