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$$\displaystyle{\frac{(1+x)^n}{(1-x)^3}=a_{0}+a_{1}x+a_{2}x^2+\cdots}$$, show that $$\displaystyle{a_{0}+\cdots+a_{n-1}=\frac{n(n+2)(n+7)2^{n-4}}{3}}$$

When i gave this problem to my friends they said No. problem looks like nonsense. The sum of the coefficients of this polynomial should be 2^(n-3). And approximations from number three dint come close to any value that would make sense.Look at n=4 in problem The polynomial you get is x+1 Sum of the coefficients is two Which is not equal to 88 How is the equality in the problem supposed to hold? Still i try to show

My work

For $\displaystyle{\left| x \right| < 1}$ we have $\displaystyle{\frac{1}{{1 - x}} = \sum\limits_{n = 0}^\infty {{x^n}} }$, and also $$\displaystyle{{\left ( {x + 1} \right)^n} = \sum\limits_{i = 0}^n {\left( {\begin{array}{*{20}{c}} n \\ i \\ \end{array} } \right){x^i}} = \sum\limits_{i = 0}^n {\left( {\begin{array}{*{20}{c}} n \\ i \\ \end{array} } \right){x^{n - i}}} }$$. Then with two productions and "playing" with the indices we have $$\displaystyle{\frac{1}{{{{\left( {1 - x} \right)}^2}}} = \sum\limits_{n = 0}^\infty {\left( {n + 1 } \right){x^n}} {\text{ & }}\boxed{\frac{1}{{{{\left( {1 - x} \right)}^3}}} = \frac {1}{2}\sum\limits_{n = 0}^\infty {\left( {n + 2} \right)\left( {n + 1} \right){x^n}} }},$$ so $$\displaystyle{\frac{{{{\left( {x + 1} \right)}^n}}}{{{{\left( {1 - x} \right)}^3}}} = \frac{1}{2}\left( {\sum\limits_{i = 0}^n {\left( {\begin{array}{*{20}{c}} n \\ i \\ \end{array} } \right){x^i}} } \right)\left( {\sum\limits_{j = 0}^\infty {\left( {j + 2} \right)\left( {j + 1} \right){x^j}} } \right)}$$ with the coefficient

$$\displaystyle{\frac{1}{2}\left( {\left( {\begin{array}{*{20}{c}} n \\ 0 \\ \end{array} } \right)\left( {k + 2} \right)\left( {k + 1} \right) + \left( {\begin{array}{*{20}{c}} n \\ 1 \\ \end{array} } \right)\left( {k + 1} \right)k + \left( {\begin{array}{*{20}{c}} n \\ 2 \\ \end{array} } \right)k\left( {k - 1} \right) + .. + \left( {\begin{array}{*{20}{c}} n \\ k \\ \end{array} } \right)2 \cdot 1} \right) = \frac{1}{2}\sum\limits_{i = 0}^k {\left( {\begin{array}{*{20}{c}} n \\ i \\ \end{array} } \right)} \left( {k + 2 - i} \right)\left( {k + 1 - i} \right)}$$

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    $\begingroup$ $2^{n-3}$ would be the answer if the numerator was $(1-x)^n$, but it is $(1+x)^n$... $\endgroup$ Dec 3, 2023 at 19:14
  • $\begingroup$ @OleksandrKulkov Now they said No. Numerator and denominator are both 1+x. The whole quotient is (1+x)^(n-3). By using the binomial theorem we see that the coefficients are only binomial coefficients nad summing them up gives a power of two. If the numerator and the denominator were 1-x the sum would be alternating which means that the result would be 0. $\endgroup$
    – user1235604
    Dec 3, 2023 at 19:52

1 Answer 1

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Multiplying with $\frac{1}{1-x}$ turns a genfunc into genfunc for its own prefix sums.

Thus, you essentially want to find

$$\begin{align} [x^{n-1}] \frac{1}{1-x} \frac{(1+x)^n}{(1-x)^3} &= [x^{n-1}] \frac{(1+x)^n}{(1-x)^{4}}. \end{align}$$

If we take $x^k$ from $\frac{1}{(1-x)^4} = \sum\limits_k \binom{k+3}{3} x^k$ and $x^{n-k-1}$ from $(1+x)^n = \sum\limits_k \binom{n}{k} x^k$, this expands into

$$\begin{align} = \sum\limits_{k=0}^{n-1} \binom{n}{k+1} \binom{k+3}{3}.\end{align}$$

Now, we substitute $\binom{k+3}{3} = [x^3](1+x)^{k+3}$ to get

$$\begin{align} &= [x^3](1+x)^2 \sum\limits_{k=1}^{n} \binom{n}{k} (1+x)^{k} \\ &= [x^3] (1+x)^2 \left((2+x)^n-1\right) \\ &= [x^3](2+x)^n + 2[x^2] (2+x)^n +[x^1](2+x)^n. \end{align}$$

Note that $(2+x)^n = 2^n + n 2^{n-1} x + \binom{n}{2} 2^{n-2} x^2 + \binom{n}{3} 2^{n-3} x^3 + O(x^4)$, hence we get

$$ = \binom{n}{3} 2^{n-3} +2\binom{n}{2} 2^{n-2} + n2^{n-1} = \frac{2^n n(n+2)(n+7)}{48}. \square $$

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  • $\begingroup$ Thnaks alot for answer(+1) $\endgroup$
    – user1235604
    Dec 3, 2023 at 19:18

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