Question

Bài 1: cho \(A=1+2+2^2+2^3+...+2^{100}\).  tìm số dư khi chia cho 31.

Bài 2: Biết : \(1^2+2^2+3^2+...+10^2=385\). tính \(S=2^2+4^2+6^2+...+20^2\)

Answer 1

bài 1: A=1+2+2²+2³+...+2¹⁰⁰

=(1+2+2²+2³+2⁴)+(2⁵+2⁶+2⁷+2⁸+2⁹)+...+(2⁹⁶+2⁹⁷+2⁹⁸+2⁹⁹+2¹⁰⁰)

=31+2⁵.(1+2+2²+2³+2⁴)+....+2⁹⁶.(1+2+2²+2³+2⁴)

=31+2⁵.31+....+2⁹⁶.31

=31.(1+2⁵+...+2⁹⁶) chia hết cho 31

Vậy số dư khi chia A cho 31 là 0

bài 2:

S=2²+4²+6²+...+20²

=1².2²+2².2²+2².3²+...+2².10²

=2².(1²+2²+3²+....+10²)

=4.385=1540

Answer 2

1)dư 1

2) S=2^2+4^2+6^2+..+20^2

=(1.2)^2+(2.2)^2+(2.3)^2+...+(2.10)^2

=1^2.2^2+2^2.2^2+2^2.3^2+..+2^2.10^2

=2^2.(1^2+2^2+3^2+...+10^2)=4.385=1540

tick nhé

Answer 3

S=2^2+4^2+6^2+...+20^2

=(1.2)^2+(2.2)^2+(3.2)^2+...+(10.2)^2

=1.2^2+2^2.2^2+3^2.2^2+...+10^2.2^2

=2^2(1+2^2+3^2+...+10^2)

=2^2.385=1540

Answer 4

A=1+2+2^2+2^3+..+2^100

=(1+2+2^2+2^3+2^4)+...+(2^96+2^97+2^98+2^99+2^100)

=31+...+2^96(1+2+2^2+2^3+2^4)

=31+...+2^96.31

=31(1+...+2^96) chia hết cho 31

=> số dư chia cho A=0