Ta chỉ cần tách các tổng thành tích thôi em nhé :)

a. \(8.2^n+2^{n+1}=8.2^n+2.2^n=10.2^n\) có tận cùng là chữ số 0.

b. \(A=27.3^n-2.3^n+32.2^n-7.2^n=25.3^n+25.2^n=25\left(3^n+2^n\right)\) nên A chia hết 25.

a, Ta có : 8.2ⁿ + 1^{n + 1} 

= 8.2ⁿ + 1 (vì 1^{n + 1} lúc nào cũng bằng 1)

= 2^{3 + n} . 1

Mà 2^{3 + n} luôn luôn ko chia hết cho10

Nên 8.2ⁿ + 1^{n + 1}  ko chi hết cho10

a = 2\(^{n+1}\)(4+1) =10.2\(^n\) tận cùng =0

b= 3\(^n\)(27 -2) + 2\(^n\)(32-7)

= 25 (3\(^n\)+2\(^n\)) chia hết cho 25

a.8.2ⁿ+2ⁿ⁺¹=2ⁿ(8+2)=2ⁿ.10 có tận cùng là 0

=>đpcm

b.3ⁿ⁺³-2.3ⁿ+2ⁿ⁺⁵-7.2ⁿ=3ⁿ(27-2)+2ⁿ(32-7)

=25.3ⁿ+25.2ⁿ=25(3ⁿ+2ⁿ) chia hết cho 25

=>đpcm

a) Ta có: \(8\times2^n+2^{n+1}\) \(=8\times2^n+2^n\times2\) \(=2^n\times\left(8+2\right)\) \(=2^n\times10\) \(=...0\)

Vậy \(8\times2^n+2^{n+1}\) có tận cùng bằng chữ số 0 (đpcm).

b) Ta có: \(3^{n+3}-2\times3^n+2^{n+5}-7\times2^n\) \(=3^n\times3^3-2\times3^n+2^n\times2^5-7\times2^n\) \(=3^n\times\left(3^3-2\right)+2^n\times\left(2^5-7\right)\) \(=3^n\times\left(27-2\right)+2^n\times\left(32-7\right)\) \(=3^n\times25+2^n\times25\) \(=\left(3^n+2^n\right)\times25\)

Vì \(25⋮25\)

nên \(\left(3^n+2^n\right)\times25⋮25\)

Vậy \(3^{n+3}-2\times3^n+2^{n+5}-7\times2^n\) chia hết cho 25 (đpcm).

1. \(A=\frac{1}{2}-\frac{2}{5}+\frac{1}{3}+\frac{5}{7}-\frac{-1}{6}+\frac{-4}{35}+\frac{1}{41}\)

\(=\frac{1}{2}-\frac{2}{5}+\frac{1}{3}+\frac{5}{7}+\frac{1}{6}-\frac{4}{35}+\frac{1}{41}\)

\(=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}\right)-\left(\frac{2}{5}-\frac{5}{7}+\frac{4}{35}\right)+\frac{1}{41}\)

\(=\left(\frac{5}{6}+\frac{1}{6}\right)-\left(\frac{-11}{35}+\frac{4}{35}\right)+\frac{1}{41}\)\(=1-\frac{-7}{35}+\frac{1}{41}=1+\frac{1}{5}+\frac{1}{41}=\frac{251}{205}\)

2. a) \(1+4+4^2+4^3+......+4^{99}=\left(1+4\right)+\left(4^2+4^3\right)+.......+\left(4^{98}+4^{99}\right)\)

\(=\left(1+4\right)+4^2\left(1+4\right)+.........+4^{98}\left(1+4\right)\)

\(=5+4^2.5+........+4^{98}.5=5\left(1+4^2+.....+4^{98}\right)⋮5\)( đpcm )

b) \(3^{n+2}-2^{n+2}+3^n-2^n=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)

\(=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)=3^n\left(9+1\right)-2^n\left(4+1\right)\)

\(=3^n.10-2^n.5=3^n.10-2^{n-1+1}.5=3^n.10-2^{n-1}.2.5\)

\(=3^n.10-2^{n-1}.10=10\left(3^n-2^{n-1}\right)⋮10\)( đpcm )