Question

If ${\alpha }_0$,${\alpha }_1$,${\alpha }_2$,....................${\alpha }_9$ be the roots of the equation $x^{10}$ -1 = 0. Find the value of ${\sum }_{i=0}^9\frac{1}{2-{\alpha }_i}$ (Neglect the decimal part of your answer write only integer part)

___

Answer 1

$x^{10}$ = 1

x = $(1{)}^{\frac{1}{10}}$

$x^{10}$ - 1 = $(x-{\alpha }_0)$ $(x-{\alpha }_1)$ $(x-{\alpha }_2)$............$(x-{\alpha }_9)$

Taking log on both sides

$\log (x^{10}-1)$ = $\log (x-{\alpha }_0)$ + $\log (x-{\alpha }_1)$ +....................$\log (x-{\alpha }_9)$

Differentiating both sides

$\frac{d}{dx}$$\log (x^{10}-1)$ = $\frac{d}{dx}$$\log (x-{\alpha }_0)$ + $\frac{d}{dx}$$\log (x-{\alpha }_1)$ +....................$\frac{d}{dx}$$\log (x-{\alpha }_9)$

$\frac{10.x^9}{x^{10}-1}$ = $\frac{1}{(x-{\alpha }_0)}$ + $\frac{1}{(x-{\alpha }_1)}$ + $\frac{1}{(x-{\alpha }_2)}$ +.....................$\frac{1}{(x-{\alpha }_9)}$-----------(1)

Put x = 2

$\frac{{10.2}^9}{2^{10}-1}$ - ${\sum }_{i=0}^9\frac{1}{2-{\alpha }_i}$

${\sum }_{i=0}^9\frac{1}{2-{\alpha }_i}$ = $\frac{10\times 512}{1024-1}$ = $\frac{10\times 512}{1023}$ = $\frac{5120}{1023}$ = 5.00488

So, correct answer = 5

Or we can find approximate integer value by eliminating 1 from the denominator so, $\frac{10\times 512}{1024-1}$ ≈ $\frac{10\times 512}{1024}$ ≈

$\frac{10\times 2^9}{2^{10}}$ = 5