Question

a)Chứng minh rằng :\(7^6+7^5-7^4⋮55\)

b)Tính \(A=1+5+5^2+...+5^{50}\)

Answer 1

a)Ta có:\(7^6+7^5-7^4=7^4\left(7^2+7-1\right)=7^4.55\)

=>\(7^6+7^5-7^4⋮55\)

b)\(A=1+5+5^2+...+5^{50}\)

\(5A=5\left(1+5+5^2+...+5^{50}\right)=5+5^2+5^3+...+5^{51}\)

\(5A-A=5+5^2+5^3+...+5^{51}-\left(1+5+5^2+...+5^{50}\right)\)

\(4A=5^{51}-1\)

\(\Rightarrow A=\dfrac{5^{51}-1}{4}\)

Answer 2

a) \(7^6+7^5+7^4=7^4\left(7^2+7+1\right)\)

= \(7^4.55\)

Vậy: \(7^6+7^5+7^4\) chia hết cho 55.

b) A= \(1+5+5^2+5^3+5^4+.....+5^{50}\)

5A= 5+\(5^2+5^3+5^4+5^{51}\)

5A-A= 5+\(5^2+5^3+5^4+......+5^{51}\)\(-\left(1+5^2+5^3+5^4+......+5^{51}\right)\)

4A= 5+\(5^2+5^3+5^4+......+5^{51}\)\(-1-5-5^2-5^3-5^4-.......-5^{50}\)

= \(5^{51}-1\)

Vậy A= \(\left(5^{51}-1\right):4\)

Tick mk nha!

Answer 3

\(7^6+7^5-7^4\)

\(=7^4.7^2+7^4.7^1-7^4.1\)

\(=7^4.49+7^4.7-7^4.1\)

\(=7^4\left(49+7-1\right)\)

\(=7^4.55⋮55\)

\(A=1+5+5^2+...+5^{50}\)

\(5A=5\left(1+5+5^2+...+5^{50}\right)\)

\(5A=5+5^2+5^3+...+5^{51}\)

\(5A-A=\left(5+5^2+5^3+...+5^{51}\right)-\left(1+5+5^2+...+5^{50}\right)\)

\(4A=5^{51}-1\)

\(A=\dfrac{5^{51}-1}{4}\)