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Now $\sum_{i=0}^{10}(x_{i}+1) L _{10,i}(5) = (x_{0}+1) L _{10,0}(5) + (x_{1}+1) L _{10,1}(5) + ... + (x_{10}+1) L _{10,10}(5)$

Which I can further decompose into

$\frac{(x_{0}+1)(5-x_{1})(5-x_{2})...(5-x_{10})}{(x_{0}-x_{1})(x_{0}-x_{2})...(x_{0}-x_{10})} + \frac{(x_{1}+1)(5-x_{0})(5-x_{2})...(5-x_{10})}{(x_{1}-x_{0})(x_{1}-x_{2})...(x_{1}-x_{10})} + ... + \frac{(x_{10}+1)(5-x_{0})(5-x_{2})...(5-x_{9})}{(x_{10}-x_{0})(x_{10}-x_{2})...(x_{10}-x_{9})}$

since we know the formula for $L _{10,i}(5)$.

I need help simplifying this expression, because I think there should be a trick somewhere to making this question much easier to solve, but I just don't know it yet. I don't think it would be wise to expand this out as there are 11 $ x_{i}$ terms which would mean 11 x 10 = 110 terms in the denominator.