Question

Chứng minh rằng 2⁰ +2¹ +2² +...+2^5n-3 +2^5n-2 +2^5n-1  chia hết cho 31 nếu n là số nguyên dương bất kif

help me please

Answer 1

Đặt A=\(2^0+2^1+2^2+....+2^{5n-3}+2^{5n-2}+2^{5n-1}\)

\(\Leftrightarrow A=\left(2^0+2^1+2^2+2^3+2^4+2^5\right)+...+\left(2^{5n+2}+2^{5n+1}+2^{5n}+2^{5n-1}+2^{5n-2}+2^{5n-3}\right)\)

\(\Leftrightarrow A=2^0\left(1+2+2^2+2^3+2^4\right)+....+2^{5n+2}\left(1+2+2^2+2^3+2^4\right)\)

\(\Leftrightarrow A=2^0\cdot31+2^5\cdot31+....+2^{5n+2}\cdot31\)

\(\Leftrightarrow A=31\cdot\left(2^0+2^5+...+2^{5n+2}\right)\)

\(\Rightarrow A⋮31\left(đpcm\right)\)

Answer 2

quynh oi dpcm la gi vay?

Answer 3

Là bạn muốn tớ thử đấy nhé!

Đặt S=1+2+2²+...+2^5n-3+2^5n-2+2^5n-1

=(1+2²+2⁴+2⁶+2⁸)+(2+2³+2⁵+2⁷+2⁹)+...+(2^5n-7+2^5n-5+2^5n-3+2^5n-1)

=1(1+2²+2⁴+2⁶+2⁸)+2(1+2²+2⁴+2⁶+2⁸)+...+2^5n-7(1+2²+2⁴+2⁶+2⁸)

=1.31+2.31+...+2^5n-7.31 chia hết cho 31

Vậy ...

Answer 4

y to la ban thu ca hai cau cau to nhan rieng va cau nay