1.

\(A=7+7^2+7^3+...+7^{78}\)

\(=\left(7+7^2\right)+\left(7^3+7^4\right)+...+\left(7^{77}+7^{78}\right)\)

\(=7\left(1+7\right)+7^3\left(1+7\right)+...+7^{77}\left(1+7\right)\)

\(=7\cdot8+7^3\cdot8+...+7^{77}\cdot8\)

\(=\left(7+7^3+...+7^{77}\right)\cdot8\) chia hết cho 8

Vậy A chia hết cho 8 (đpcm)

\(A=3+3^2+3^3+...+3^{155}\)

\(=\left(3+3^2+3^3+3^4+3^5\right)+...+\left(3^{151}+3^{152}+3^{153}+3^{154}+3^{155}\right)\)

\(=3\left(1+3+3^2+3^3+3^4\right)+...+3^{151}\left(1+3+3^2+3^3+3^4\right)\)

\(=\left(3+...+3^{151}\right)\cdot121\) chia hết cho 121

Vậy A chia hết cho 121 (đpcm)

\(3+3^2+3^3+...+3^{2012}\)

\(=\left(3+3^2+3^3+3^4\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)

\(=3\left(1+3+3^2+3^3\right)+...+3^{2009}\left(1+3+3^2+3^3\right)\)

\(=40\left(3+...+3^{2009}\right)⋮40\)

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rrrrr

\(S=1+2+2^2+...+2^{99}\)

\(S=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{98}+2^{99}\right)\)

\(S=3+2^2.3+...+2^{98}.3\)

\(=3\left(1+2^2+...+2^{98}\right)⋮3\)

Giải giùm tớ (-209)-401+12

a,n-3 chia hết n+3

có n-3 chia hết n+3

<=> n+3-6chia hết n+3

vì n+3 chia hết n+3 nên 6 chia hết n+3

=>n+3 thuộc ước 6 ={1;2;3;6}

=> n = 4;5;6;9

\(3+\frac{3}{5}+\frac{3}{25}+\frac{3}{125}+\frac{3}{625}\)

\(=3\left(1+\frac{1}{5}+\frac{1}{25}+\frac{1}{125}+\frac{1}{625}\right)\)

\(=3\cdot\frac{781}{625}\)

\(=\frac{2343}{625}\)

Cảm ơn bạn rất rất nhìu nhé.

Đặt A = \(3+\frac{3}{5}+\frac{3}{25}+\frac{3}{125}\)\(+\frac{3}{625}\)

=> 625 x A = 625 x 3 + 625 x\(\frac{3}{5}\)+ 625 x \(\frac{3}{25}\)+ 625 x \(\frac{3}{125}\)+ 625 x \(\frac{3}{625}\)

=> 625 x A = 1875 + 375 + 75 + 15 + 3

                  = 2343

=> A = \(\frac{2343}{625}\)

Lời giải:

$C=1+3^2+3^3+(3^4+3^5+3^6)+(3^7+3^8+3^9)+....+(3^{97}+3^{98}+3^{99})$

$=37+3^4(1+3+3^2)+3^7(1+3+3^2)+...+3^{97}(1+3+3^2)$

$=11+13.2+(1+3+3^2)(3^4+3^7+...+3^{97})$
$=11+13.2+13(3^4+3^7+...+3^{97})$

$=11+13(2+3^4+3^7+....+3^{97})$

$\Rightarrow C$ chia $13$ dư $11$.