Simplify and write the answer with positive integral exponents (a) (2⁵÷2¹¹)²×2–³
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Answer:
There is really no need for a long answer. Visual approach.
The representation of xnxn in basexbasex is always a one followed by nn zeroes. So if we add all of them starting at 00 we get n+1n+1 ones in basexbasex.
If we dont add the first (20)(20), as in your question, we only get n ones followed by one zero. Thats 12 ones and a zero, which is 81908190 in decimal.
Special case in base2base2:
Adding 1 to a number that only consists of the highest digit (x−1(x−1inbasexbasex) all digits will be zero, lead by a one.
In the special case for base 2 the one is the highest digit, therefore adding all and then adding 1 will give us the next power: 2(n+1)2(n+1)
sum(2i,0,n)+1=2(n+1)sum(2i,0,n)+1=2(n+1)
For n=12n=12 we get
213−1=8191213−1=8191
Since you asked for the first (20=1)(20=1) term beeing removed we subtract that and get 81908190.
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George Coote, MMath Mathematics, University of Warwick (2023)
Answered May 29, 2018 · Upvoted by Richard Enison, M.A. Mathematics & Mathematical Logic and Foundations, University of California, Berkeley (1974) · Author has 412 answers and 666.8K answer views