I have to find coefficient of $x^3$ in $(x + 1)^n + (x + 1)^{n - 1}(x + 2) + (x + 1)^{n - 2}(x + 2)^2 + \ldots + (x + 2)^n$ and I cannot get a starting point as to how to solve this.
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2Hint: $a^{n+1}-b^{n+1}=(a-b)(a^n+a^{n-1}b+\cdots+ab^{n-1}+b^n)$. Call $a=x+2$ and $b=x+1$. – DiegoMath Jun 10 '21 at 00:46
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@DiegoMath Thanks my brain has frozen at 5AM. – Shiv Jun 10 '21 at 01:08
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We can start by simplifying the sum as follows but also note that we assume here $n\ge 3$, $$ \displaystyle \sum_{k=0}^n (x+1)^{n-k}(x+2)^{k} = (x+2)^{n+1}-(x+1)^{n+1}$$ This can be derived with the help of finite geometric series. Now the coefficient of $x^r$ in $(x+a)^p$ is, $$\displaystyle [x^r](a+x)^p = \binom{p}{r}a^{p-r}$$ Using this the coefficient of $x^3$ is, $$ \mathfrak{L} = \binom{n+1}{3}\left(2^{n-2}-1\right)$$ This is the required answer.
Aditya Narayan Sharma
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