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Simplify $$0^2 \binom{n}{0} + 1^2 \binom{n}{1} + 2^2 \binom{n}{2} + \dots + n^2 \binom{n}{n},$$
where $$n ≥ 2$$.

Enter your answer in the form $$f(n)2^{g(n)}$$, where $$f(n)$$ and $$g(n)$$ are polynomials in $$n$$ with integer coefficients.

Aug 7, 2020

#1
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Simplify

$$0^2 \dbinom{n}{0} + 1^2 \dbinom{n}{1} + 2^2 \dbinom{n}{2} + 2^3 \dbinom{n}{2} + \dots + (n-1)^2 \dbinom{n}{n-1}+ n^2 \dbinom{n}{n}$$,

where $$n \ge 2$$.

$$\begin{array}{|rcll|} \hline \mathbf{S} &=& \mathbf{0^2 \dbinom{n}{0} + 1^2 \dbinom{n}{1} + 2^2 \dbinom{n}{2} + 2^3 \dbinom{n}{2} + \dots + (n-1)^2 \dbinom{n}{n-1}+ n^2 \dbinom{n}{n} } \\\\ \mathbf{S} &=& \mathbf{\sum \limits_{k=0}^{n} k^2\dbinom{n}{k} } \\\\ S &=& \sum \limits_{k=0}^{n} k*k*\dbinom{n}{k} \quad | \quad \mathbf{k = \dbinom{k}{1}} \\\\ S &=& \sum \limits_{k=0}^{n} k\dbinom{k}{1}\dbinom{n}{k} \quad | \quad \color{red}\dbinom{k}{1}\dbinom{n}{k}= \dbinom{n}{1}\dbinom{n-1}{k-1} \\\\ S &=& \sum \limits_{k=1}^{n} k\dbinom{n}{1}\dbinom{n-1}{k-1} \quad | \quad \mathbf{\dbinom{n}{1}=n } \\\\ S &=& \sum \limits_{k=1}^{n} kn\dbinom{n-1}{k-1} \\\\ \mathbf{S} &=& \mathbf{ n\sum \limits_{k=1}^{n} k\dbinom{n-1}{k-1} }\\ \hline \end{array}$$

$$\mathbf{ \sum \limits_{k=1}^{n} k\dbinom{n-1}{k-1} = \ ? }$$

$$\begin{array}{|rcll|} \hline \mathbf{\left( 1+x \right)^{n-1}} &=& \mathbf{\dbinom{n-1}{0}x^0 +\dbinom{n-1}{1}x^1 + \dbinom{n-1}{2}x^2 + \dots + \dbinom{n-1}{n-1}x^{n-1}} \quad | \quad *x \\\\ x\left( 1+x \right)^{n-1} &=& \dbinom{n-1}{0}x^1 +\dbinom{n-1}{1}x^2 + \dbinom{n-1}{2}x^3 + \dots + \dbinom{n-1}{n-1}x^{n} \\\\ x\left( 1+x \right)^{n-1} &=& \sum \limits_{k=0}^{n-1} \dbinom{n-1}{k}x^{k+1} \\\\ \dfrac{d}{dx}x\left( 1+x \right)^{n-1} &=& \sum \limits_{k=0}^{n-1} (k+1) \dbinom{n-1}{k}x^{k} \\\\ x(n-1) \left(1+x \right)^{n-2} + \left( 1+x \right)^{n-1} &=& \sum \limits_{k=0}^{n-1} (k+1) \dbinom{n-1}{k}x^{k} \quad | \quad \mathbf{x=1} \\\\ (n-1) \left(1+1 \right)^{n-2} + \left( 1+1 \right)^{n-1} &=& \sum \limits_{k=0}^{n-1} (k+1) \dbinom{n-1}{k}1^{k} \\\\ (n-1) \left(2 \right)^{n-2} + \left( 2 \right)^{n-1} &=& \sum \limits_{k=0}^{n-1} (k+1) \dbinom{n-1}{k} \\\\ (n-1) 2^{n-2} + 2^{n-1} &=& \mathbf{\sum \limits_{k=1}^{n} k \dbinom{n-1}{k-1}} \\ \hline \end{array}$$

$$\mathbf{S= \ ?}$$

$$\begin{array}{|rcll|} \hline \mathbf{S} &=& \mathbf{ n\sum \limits_{k=1}^{n} k\dbinom{n-1}{k-1} } \quad | \quad \mathbf{\sum \limits_{k=1}^{n} k \dbinom{n-1}{k-1} = (n-1) 2^{n-2} + 2^{n-1} }\\\\ S &=& n\left( (n-1) 2^{n-2} + 2^{n-1} \right) \\\\ S &=& n\left( (n-1) 2^{n-2} + 2^{n-1}*\dfrac{2}{2} \right) \\\\ S &=& n\left( (n-1) 2^{n-2} + 2^{n-2}*2 \right) \\\\ S &=& n2^{n-2}\left( n-1 + 2 \right) \\\\ S &=& n2^{n-2}\left( n+1 \right) \\\\ \mathbf{S} &=& \mathbf{n(n+1)2^{n-2}} \quad | \quad f(n)= n(n+1),\ g(n) = n-2,\ n \ge 2\\ \hline \end{array}$$

Aug 10, 2020
edited by heureka  Aug 10, 2020
edited by heureka  Aug 10, 2020