Question

Cho bieu thuc 

\(a=1+7+7^2+7^3+...+7^{100}\)

a)  Tinh a 

b) Tim x biet \(6a+1=7^x\)

c) Tim du trong phep chia a cho 8

Answer 1

c)\(a=1+7+7^2+...+7^{100}\)

\(a=\left(1+7\right)+\left(7^2+7^3\right)+...+\left(7^{99}+7^{100}\right)\)

\(a=8+7^2\left(1+7\right)+...+7^{99}\left(1+7\right)\)

\(a=8+7^2.8+...+7^{99}.8\)

\(a=8\left(1+7^2+...+7^{99}\right)\) chia hết cho 8

=> a chia 8 dư 0

Answer 2

a)\(a=\left(1+7+7^2+...+7^{100}\right)\)

=>\(7a=7\left(1+7+7^2+...+7^{100}\right)=7+7^2+7^3+...+7^{101}\)

=>\(7a-a=\left(7+7^2+7^3+...+7^{101}\right)\)\(-\left(1+7+7^2+...+7^{100}\right)\)

=>\(6a=7^{101}-1\Rightarrow a=\frac{7^{101}-1}{6}\)

b)\(6a+1=7^x\Rightarrow6.\frac{7^{101}-1}{6}+1=7^x\Rightarrow7^{101}-1+1=7^x\Rightarrow7^{101}=7^x\)=>x=101

Answer 3

Cảm ơn cậu nka