If N>1, there are (2N-1) - (2N-1-1), otherwise, 1 nodes in the Nth level of a balanced binary tree.
Level N of a binary tree has, at most, 2^N nodes. Note that the root node is regarded as being level 0. If we regard it as being level 1, then level N would have 2^(N-1) nodes at most.
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2^6 = 64 haploid gametes use of 2^nth rule.
The Nusselt number is proportional to the Prandtl number to the nth power, where n is a positive number less than one.
tn = t1+(n-1)d -- for arithmetic tn = t1rn-1 -- for geometric
Level N of a binary tree has, at most, 2^N nodes. Note that the root node is regarded as being level 0. If we regard it as being level 1, then level N would have 2^(N-1) nodes at most.
2(n-1)
lea-nth. Formulate your question please
Yes. You can have as many nth terms as you can be bothered to write down!
The number of odd numbers in the Nth row of Pascal's triangle is equal to 2^n, where n is the number of 1's in the binary form of the N. In this case, 100 in binary is 1100100, so there are 8 odd numbers in the 100th row of Pascal's triangle.
nth n = 1,2,3,.......
If a set has N elements then it has 2N subsets. So you can see that a list of all subsets soon becomes a very big task. For reasonably small values of N, one way to generate all subsets is to list the binary numbers from 0 to 2N. Then, each of these represents a subset of the original set. If the nth digit is 0 then the nth element is not in the set and if the nth digit is 1 then the nth element is in the set. That will generate all the subsets.
Well, isn't that just a lovely pattern we have here? Each term is increasing by 4, isn't that delightful? So, if we want to find the nth term, we can use the formula: nth term = first term + (n-1) * common difference. Just like painting a happy little tree, we can plug in the values and find the nth term with ease.
One of the infinitely many possible rules for the nth term of the sequence is t(n) = 4n - 1
The Nth Bit Labs
Nth Man was created in 1976.
3^n