S= 1^3+2^3+3^3+...+100^3                                                                                                                             S=1^2*1+2^2*2+3^2*3+...+100^2*100                                                                                                             S=(100*101*201)/6+5050                                                                                                                               S=5126002500

ta có \(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+3}+2^{n+2}\)

        \(\Rightarrow3k+3k+2k+2k+2k\)

        \(=k\left(3+3+2+2+2\right)\)

     \(=12k=6.2k\) chia hết cho 6

=<2014-2013><2013+2014>+<2012-2011><2012+2011>...<2-1><2+1>                                                                                                       =4027+ 4023 +...+3

=<4027 -3>:4 +1 =1007

Giải rồi trả lời cái j nữa 

Bó tay

Mình đã giải xong, không cần nữa nha! Thanks!

Bài 1 : 

A=2+22+23+...+299+2100A=2+22+23+...+299+2100

⇒2A=22+23+24+...+2100+2101⇒2A=22+23+24+...+2100+2101

⇒A=2101−2⇒A=2101−2

B=3+32+33+...+399+3100B=3+32+33+...+399+3100

⇒3B=32+33+34+...+3100+3101⇒3B=32+33+34+...+3100+3101

Bài 2 :

2.Chứng minh rằng

2¹²+3¹²+2¹³+2¹⁴+3¹⁵ chia hết cho 7

⇒2B=3101−3⇒2B=3101−3

⇒B=3101−32

Ok tối mk giẳi cho