Question

1+2+4+8+16+...=-1 ??!!

Answer 1

1+2+4+8+16+... =(2－1)(1+2+4+8+16+...) =2(1+2+4+8+16+...)－(1+2+4+8+16+...) =2+4+8+16+32+...－1－2－4－8－16－... =－1 Sol 1+2+4+8+16+... =(2－1)(1+2+4+8+16+...) =>正确 =2(1+2+4+8+16+...)－(1+2+4+8+16+...) =>错误 因为(1+2+4+8+16+…) 趋近于 ∞ 我们不能写成 ∞－∞
Let x be the last number. 1+2+4+8+16+...+x =(2-1)(1+2+4+8+16+...+x =2(1+2+4+8+16+...+x)-1(1+2+4+8+16+...+x) =(2+4+8+16+32+...+2x)-1-2-4-8-16-...-x =x-1 As the result
we know that the wer should be (the last number -1). But why? It's because many people fet to use 2 times the last munber.Therefore
they will get (-1) as the wer after they finish it.
参考: myself
上面岩 首先 如果1个1个term咁睇 2-1 =1 / 4-2=2 咁仍然都系1+2+4+8+16+... 其次 而上面讲紧既就系 前面你有n咁多个term 后面减既个部分又有n咁多个term 如果你抽左-1 出尼 咁咪会变成n terms : n-1 terms 咁前面多左既term就会系抵消番d illusion 无论系边个方法睇 obviously 唔系=-1 !
1+2+4+8+... should be 1+2+4+8+...+2^n in your calculation 1+2+4+8+...+2^n =(2-1)(1+2+4+8+...+2^n) =2+4+8+16+...+2^(n+1)-1+2+4+8+...+2^n the wer is not -1!
很明显，no. of terms of (2+4+8+16+32+...) = no. of terms of (-1-2-4-8-16-...) 所以只以 (2+4+8+16+32+...) 抵销 (-1-2-4-8-16...) 的话 (2+4+8+...) 最后的1个term 会出了多来~_~