Nhóm các hạng tử của tổng đã cho theo dạng sau:

\(A=\left(7+7^2+7^3+7^4\right)+\left(7^5+7^6+7^7+7^8\right)+...+\left(7^{4k-3}+7^{4k-2}+7^{4k-1}+7^{4k}\right)\)

     \(=\left(7+7^2+7^3+7^4\right)+7^4\left(7+7^2+7^3+7^4\right)+...+7^{4k-4}\left(7+7^2+7^3+7^4\right)\)

     \(=\left(7+7^2+7^3+7^4\right)\left(1+7^4+7^8+...+7^{4k-4}\right)\)

     \(=7\left(1+7+7^2+7^3\right)\left(1+7^4+7^8+...+7^{4k-4}\right)\)

\(A=7\left(1+7+49+343\right)\left(1+7^4+7^8+...+7^{4k-4}\right)=7.400.B\)

Vậy,   \(A\)  chia hết cho  \(400\)

Ta có : \(A=7+7^2+7^3+...+7^{4k}\)

\(=\left(7+7^2+7^3+7^4\right)+...+\left(7^{4k-3}+7^{4k-2}+7^{4k-1}+7^{4k}\right)\)

\(=\left(7+7^2+7^3+7^4\right)+...+7^{4k-4}\left(7+7^2+7^3+7^4\right)\)

\(=\left(7+7^2+7^3+7^4\right)\left(1+...+7^{4k-4}\right)\)

\(=2800\left(1+...+7^{4k-4}\right)\)

\(=350.8\left(1+...+7^{4k-4}\right)⋮8\)

\(\Rightarrow A⋮8\left(1\right)\)

Ta lại có : \(A=7+7^2+7^3+...+7^{4k}\)

\(\Rightarrow7A=7^2+7^3+7^4+...+7^{4k+1}\)

\(\Rightarrow7A-A=\left(7^2+7^3+7^4+...+7^{4k+1}\right)-\left(7+7^2+7^3+....+7^{4k}\right)\)

hay \(6A=7^{4k+1}-7=7\left(7^{4k}-1\right)\)

Vì \(7\equiv2\left(mod5\right)\)\(\Rightarrow7^{4k}\equiv2^{4k}=16^k\left(mod5\right)\)

mà \(16\equiv1\left(mod5\right)\)\(\Rightarrow16^k\equiv1^k=1\left(mod5\right)\)

\(\Rightarrow7^{4k}\equiv1\left(mod5\right)\)

\(\Rightarrow7^{4k}-1⋮5\left(\cdot\right)\)

\(\Rightarrow7\left(7^{4k}-1\right)⋮5\)

\(\Rightarrow6A⋮5\)

Nhưng \(\left(6;5\right)=1\)

\(\Rightarrow A⋮5\left(2\right)\)

Ta lại có tiếp : \(7\equiv1\left(mod2\right)\)

\(\Rightarrow7^{4k}\equiv1^{4k}=1\left(mod2\right)\)

\(\Rightarrow7^{4k}-1⋮2\left(\cdot\cdot\right)\)

Từ \(\left(\cdot\right)\), \(\left(\cdot\cdot\right)\) và \(\left(2;5\right)=1\): \(\Rightarrow7^{4k}-1⋮10\)

\(\Rightarrow7\left(7^{4k}-1\right)⋮10\)

\(\Rightarrow6A⋮10\)

Nhưng \(\left(6;10\right)=1\)

\(\Rightarrow A⋮10\left(3\right)\)

Từ \(\left(1\right),\left(2\right),\left(3\right)\)và \(\left(5;8;10\right)=1\)

\(\Rightarrow A⋮400\left(đpcm\right)\)

7 + 7² + 7³ + ... + 7^4k

= [7 + 7² + 7³ + 7⁴] + 7⁴[7 + 7² + 7³ + 7⁴] + .... + 7^4k-4[7 + 7² + 7³ + 7⁴]

= 2800 + 7⁴.2800 + .... + 7^4k-4. 2800

= 7.400 [7⁰ + 7⁴ + ... + 7^4k-4] \(⋮400\)

mình ko bít làm

\(A=7^1+7^2+7^3+7^4+...+7^{4k}\)

\(=\left(7^1+7^2+7^3+7^4\right)+...+\left(7^{4k-3}+7^{4k-2}+7^{4k-1}+7^{4k}\right)\)

\(=7.\left(1+7+7^2+7^3\right)+...+7^{4k-3}.\left(1+7+7^2+7^3\right)\)

\(=7.\left(1+7+49+343\right)+...+7^{4k-3}.\left(1+7+49+343\right)\)

\(=7.400+...+7^{4k-3}.400=400.\left(7+...+7^{4k-3}\right)\)

\(=100.\left[4.\left(7+...+7^{4k-3}\right)\right]⋮100\)

=> đpcm