计算4+67+697+6997+...699997S=1+3/2+5/2X2+...21/2的10次方计算这两道题(简便方法)

Question

计算 4+67+697+6997+...699997
S=1+3/2+5/2X2+...21/2的10次方
计算这两道题(简便方法)

Answer 1

1.
4+67+697+6997+69997+699997
=4+(70-3)+(700-3)+(7000-3)+(70000-3)+(700000-3)
=4+777770-3*5
=777770-11
=777759
2.
2*S=2+3+5/2+7/(2^2)+...+21/(2^9)
S=2S-S
=2+3+5/2+7/(2^2)+...+21/(2^9)-[1+3/2+5/(2^2)+...+19/(2^9)+21/(2^10)]
=5-1+(5/2-3/2)+[7/(2^2)-5/(2^2)]+...+[21/(2^9)-19/(2^9]-21/(2^10)
=4+2/2+2/(2^2)+...+2/(2^9)-21/(2^10)
=4+1+1/2+...+1/(2^8)-21/(2^10)
=5+1-1/(2^8)-21/(2^10)
=6-4/(2^10)-21/(2^10)
=6-25/(2^10)
2^N,表示2的N次方.方法是这样,不过没检查步骤对不对.