Collection of Maths Problems

Coefficients

Task number: 3870

• Variant

  by \(x^{15}\) in \((x^2+x^3+x^4+\ldots)^4\)

• Hint

  Find a combinatorial argument first, then apply the generating functions and the generalized binomial theorem.

• Solution

  Combinatorial argument: we distribute 15 balls into four boxes, each shall have at least two. This is the same as if we put two in each box and then distribute the remaining seven.

  The solution using generating functions:

  \(\displaystyle (x^2+x^3+x^4+\ldots)^4= x^8(1+x+x^2+\ldots)^4= x^8\frac1{(1-x)^4}= x^8\sum_{k=0}^\infty \binom{-4}{k}(-x)^k=\\ x^8\left(… +\binom{-4}{7}(-x)^7+…\right)= … +\binom{10}{7} x^{15}+… \)

• Answer

  Coefficient by \(x^{15}\) is \(\binom{10}{3}\).

• Variant

  by \(x^5\) in \((1-2x)^{-2}\)

• Hint

  Use the generalized binomial theorem.

• Solution

  \(\displaystyle (1-2x)^{-2}= \sum_{k=0}^\infty \binom{-2}k (-2x)^k= … + \binom{-2}5(-2x)^5+ … \)

  while \(\displaystyle \binom{-2}5(-2)^5 = \frac{(-2)(-3)(-4)(-5)(-6)}{5!}\cdot(-2)^5=6 {\cdot} 2^5 \)

• Answer

  Coefficient by \(x^{5}\) is \(192\).

• Variant

  by \(x^4\) in \((1+x)^{1/3}\)

• Solution

  \( \displaystyle \binom{1/3}{4}= \frac{\frac{1}{3}\cdot\frac{-2}{3}\cdot\frac{-5}{3}\cdot\frac{-8}{3}}{4!}= -\frac{10}{3^5} \)

• Answer

  Coefficient by \(x^4\) is \(-\frac{10}{243}\).

• Variant

  by \(x^k\) in \(1\over\sqrt{1-2x}\), for any \(k \ge 0\).

• Solution

  \( \displaystyle (1-2x)^{-1/2}= \sum_{k=0}^\infty\binom{-1/2}{k}(-2)^k x^k = \sum_{k=0}^\infty \frac{1{\cdot} 3 … (2k-1)}{k!} x^k \)

  Both the numerator and denominator of the fraction can be extended by the product of even numbers in the form \(2^k k!\). We get:

  \( \displaystyle \frac{1{\cdot} 3 … (2k-1)}{k!}=\frac{1{\cdot} 3 … (2k)!}{k!\, k! \,2^k} = \binom{2k}{k}2^{-k} \)

• Answer

  Coefficient by \(x^k\) is \(\binom{2k}{k}2^{-k}\).

• Variant

  by \(x^4\) in \((x^2-5x+6)^{-1}\).

• Hint

  Decompose the rational polynomial into partial fractions.

• Solution

  \( \displaystyle \frac{1}{x^2-5x+6}= \frac{1}{(x-2)(x-3)}= \frac{1}{x-3}-\frac{1}{x-2}= -\frac{1}{3}\cdot\frac{1}{1-\frac{1}{3}x}+\frac{1}{2}\cdot\frac{1}{1-\frac{1}{2}x}= -\frac13\sum_{i=0}^\infty\left(\frac13x\right)^i +\frac12\sum_{i=0}^\infty\left(\frac12x\right)^i = \sum_{i=0}^\infty (2^{-i-1}-3^{-i-1})x^i \)

• Answer

  Coefficient by \(x^4\) is \(2^{-5}-3^{-5}=\frac{221}{7776}\).