I'm in urgent need of help solving this proof. I have my final exam tomorrow, and me passing depends on me being able to understand how to prove this statement, and similar statements. Can anyone explain to me how this is proved? $$\displaystyle \sum_{i = 0}^n {n \choose i}{i^2}= (n+1)(n){2^{n-2}}$$ for n$\ge0$ |
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We know the nbinomial expansion $$f(x):=\sum_{i=0}^n {n\choose i}x^i = (1+x)^n.$$ Note that taking the derivative and multiplying by $x$ turns $x^i$ into $ix^i$, thus $$xf'(x)=\sum_{i=0}^n {n\choose i}ix^i$$ and repeating this step $$x (f'(x)+xf''(x))=\sum_{i=0}^n {n\choose i}i^2x^i$$ We want this evaluated at at $x=1$. With $f'(x) = n(1+x)^{n-1}$ and $f''(x)=n(n-1)(1+x)^{n-2}$, we obtain $$ 1\cdot(n2^{n-1}+1\cdot n(n-1)2^{n-2})=n(n-1+2)2^{n-2}=n(n+1)2^{n-2}.$$ |
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