The final number after $999$ operations.

I wanted to know, let the numbers $1,\frac12,\frac13,\dots,\frac1{1000}$ be written on a blackboard. One may delete two arbitrary numbers $a$ and $b$ and write $a+b+ab$ instead. After $999$ such operations only one number is left. What is this final number.

I tried, let $*$ be an operation $a*b = (1+a)(1+b) - 1$, and $(a*b)*c = (1+a)(1+b)(1+c) -1$. by induction we can show that for $N$ such no.s we have $(1+a)(1+b)(1+c)\dots(1+d) - 1$, where $n\{a,b,c,\dots,d\} = N$. In the question we have numbers $1,\frac12,\frac13,\dots,\frac1{1000}$. plugging in we get the final number $1000$, but the answer given is $100$.

What am i doing wrong?

Any help appreciated.

Thanks.

Solution 1:

Your answer is correct. We have $$ 1*(1/2)*(1/3)*\cdots*(1/1000)=\frac21\cdot\frac32\cdots\frac{1001}{1000}-1=1001-1=1000. $$