Ta có; B= 3100-399+398-397+...+32-3+1
= (3100-399) + (398-397) + ... + (32-3) + 1
= 399 + 397 + ......... + 3 + 1
=> 3B = 3100 + 399 + 397 + ......... + 3
3B - B = 3100 - 1
=> B = \(\frac{3^{100}-1}{2}\)
Ta có; B= 3100-399+398-397+...+32-3+1
= (3100-399) + (398-397) + ... + (32-3) + 1
= 399 + 397 + ......... + 3 + 1
=> 3B = 3100 + 399 + 397 + ......... + 3
3B - B = 3100 - 1
=> B = \(\frac{3^{100}-1}{3}\)
B=3100-399+398-397+...+32-3+1
3B=3101-3100+399-398+...+33-32+3
3B+B=3101+1
4B=3101+1
B=\(\frac{3^{101}+1}{4}\)
