1708Shubhodip Brilliant Solution.
$\textbf{If $\mathbf{x_{1},x_{2},x_{3},.......,x_{n}}$ are the roots of the equation $\mathbf{x^n+2x^{n-1}+3x^{n-2}+.........+nx+n+1=0}$.\\\\ Then Calcuate value of $\mathbf{\sum_{k=1}^{n}\frac{x_{k}^{n+1}-1}{x_{k}-1}=}$ }
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3 Answers
211) x= 0 and 1 are not roots
2)
f(x)= x^n + 2 x^{n-1} + ...+ nx + n+1
\frac{f(x)}{x} = x^{n-1} + 2x^{n-2} +....+ n + \frac{n+1}{x}
Thus f(x)- \frac{f(x)}{x} = \frac{x^{n+1}-1}{x-1} - \frac{n+1}{x} ________(1)
3) If x_{k},\; k\; = \left\{1,2,3..n \right\} are roots of f(x) = 0 we have from (1)
\sum_{k=1}^{n}{}\frac{x_{k}^{n+1}-1}{x_{k}-1} = \sum_{k=1}^{n}{}\frac{n+1}{x_{k}} = -n
Note that \sum_{k=1}^{n}{}\frac{1}{x_{k}}= \; -\frac{n}{n+1}