Problems

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Calculate the following sums:

a) \(\binom{5}{0} + 2\binom{5}{1} + 2^2\binom{5}{2} + \dots +2^5\binom{5}{5}\);

b) \(\binom{n}{0} - \binom{n}{1} + \dots + (-1)^n\binom{n}{n}\);

c) \(\binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n}\).

Here is a fragment of the table, which is called the Leibniz triangle. Its properties are “analogous in the sense of the opposite” to the properties of Pascal’s triangle. The numbers on the boundary of the triangle are the inverses of consecutive natural numbers. Each number is equal to the sum of two numbers below it. Find the formula that connects the numbers from Pascal’s and Leibniz triangles.

Prove that for a real positive \(\alpha\) and a positive integer \(d\), \(\lfloor \alpha / d\rfloor = \lfloor \lfloor \alpha\rfloor / d\rfloor\) is always satisfied.

Author: A. Khrabrov

Do there exist integers \(a\) and \(b\) such that

a) the equation \(x^2 + ax + b = 0\) does not have roots, and the equation \(\lfloor x^2\rfloor + ax + b = 0\) does have roots?

b) the equation \(x^2 + 2ax + b = 0\) does not have roots, and the equation \(\lfloor x^2\rfloor + 2ax + b = 0\) does have roots?

Note that here, square brackets represent integers and curly brackets represent non-integer values or 0.

The segment \(OA\) is given. From the end of the segment \(A\) there are 5 segments \(AB_1, AB_2, AB_3, AB_4, AB_5\). From each point \(B_i\) there can be five more new segments or not a single new segment, etc. Can the number of free ends of the constructed segments be 1001? By the free end of a segment we mean a point belonging to only one segment (except point \(O\)).