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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=25.3\left(4^{1975}+4^{1974}+...+4^2+4+1\right)+25\)
\(=25\left(4-1\right)\left(4^{1975}+4^{1974}+...+4^2+4+1\right)+25\)
Áp dụng hằng đẳng thức, ta có : \(A=25\left(4^{1976}-1\right)+25=25.4^{1976}\)
Vậy \(A⋮4^{1976}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
đặt B = 42015 + 42014 + 42013 + ... + 42
4B = 42016 + 42015 + 42014 + ... + 43
4B - B = ( 42016 + 42015 + 42014 + ... + 43 ) - ( 42015 + 42014 + 42013 + ... + 42 )
3B = 42016 - 42
\(\Rightarrow\)B = \(\frac{4^{2016}-4^2}{3}\)hay B = \(\frac{4^{2016}-16}{3}\)
\(\Rightarrow\)A = 75 . ( \(\frac{4^{2016}-16}{3}\)+ 5 ) + 25
A = 75 . ( \(\frac{4^{2016}-16}{3}\)+ \(\frac{15}{3}\)) + 25
A = 75 . ( \(\frac{4^{2016}-1}{3}\)) + 25
A = 25 . ( 3 . \(\frac{4^{2016}-1}{3}\)) + 25
A = 25 . ( 42016 - 1 ) + 25
A = 25 . ( 42016 - 1 + 1 )
A = 25 . 42016 \(⋮\)42016
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=\left(4n+3\right)^2-5^2=\left(4n+3-5\right)\left(4n+3+5\right)=\left(4n-2\right)\left(4n+8\right)\)
\(=8\left(n-1\right)\left(n+2\right)\). Vì A chứa thừa số 8 nên A chia hết cho 8
b) \(B=\left(2n+3\right)^2-3^2=\left(2n+3-3\right)\left(2n+3+3\right)=2n\left(2n+6\right)=4n\left(n+3\right)\)
Vì B chứa thừa số 4 nên B chia hết cho 4
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=85^2-45^2+75^2-35^2+65^2-25^2+55^2-15^2\)
\(A=\left(85-45\right)\left(85+45\right)+....+\left(55-15\right)\left(55+15\right)\)
\(A=40.130+40.110+40.90+40.70\)
\(A=40.\left(130+110+90+70\right)=40.400=16000\)
b) \(B=\left(1-2\right)\left(1+2\right)+\left(3-4\right)\left(3+4\right)+...+\left(2011-2012\right)\left(2011+2012\right)\)
\(B=-3-7-11-...-4023\)
\(B=-\left(3+7+11+...+4023\right)\)
\(B=-\dfrac{\left(3+4023\right)\left[\dfrac{\left(4023-3\right)}{4}+1\right]}{2}=2025078\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : ( 2n + 5 )2 - 25 = 4n2 + 4n + 25 - 25
= 4n . ( n + 1 )
Mà 4n . ( n + 1 ) chia hết cho 4
Hay ( 2n + 5 )2 - 25 chia hết cho 4 ( ĐPCM )
Đặt \(B=4^{1975}+4^{1974}+...+4^2\)
\(\Rightarrow4B=4^{1976}+4^{1975}+...+4^3\)
\(\Rightarrow4B-B=\left(4^{1976}+4^{1975}+...+4^3\right)-\left(4^{1975}+4^{1974}+...+4^2\right)\)
hay \(3B=4^{1976}-4^2\)
\(\Rightarrow B=\frac{4^{1976}-4^2}{3}\)
\(\Rightarrow A=75\left(B+5\right)+25\)
\(=75\left(\frac{4^{1976}-4^2}{3}+5\right)+25\)
\(=25.\left(4^{1976}-16\right)+375+25\)
\(=25.4^{1976}-400+400\)
\(=25.4^{1976}⋮4^{1976}\left(đpcm\right)\)