Question

(xb-c)a. (xc-a)b.(xa-b)c= 1

Answer 1

$\large\underline{\sf{Solution-}}$

Given summation is

$\rm :\longmapsto\:\displaystyle \sf { \sum \limits^{n} _{k = 1} \binom{n}{k} \frac{( - 1) ^{k - 1} }{k} }$

can also be rewritten as

$\rm \:  =  \: \displaystyle \sf { \sum \limits^{n} _{k = 1} \: ^nC_{k} \frac{( - 1) ^{k - 1} }{k} }$

can be further open as

$\rm \:  =  \: ^nC_{1} - \dfrac{^nC_{2}}{2} + \dfrac{^nC_{3}}{3} + - - - - + \dfrac{ {( - 1)}^{n - 1} \: ^nC_{n}}{n}$

can also be rewritten as

$\rm \:  =  \: C_{1} - \dfrac{C_{2}}{2} + \dfrac{C_{3}}{3} + - - - - + \dfrac{ {( - 1)}^{n - 1} \: C_{n}}{n}$

So, it means

$\rm \:\displaystyle \sf { \sum \limits^{n} _{k = 1} \binom{n}{k} \frac{( - 1) ^{k - 1} }{k} }  =  \: C_{1} - \dfrac{C_{2}}{2} + \dfrac{C_{3}}{3} + - - - + \dfrac{ {( - 1)}^{n - 1} \: C_{n}}{n}$

Now, we know that

$\rm :\longmapsto\: {(1 - x)}^{n} = C_{0} - C_{1}x + C_{2} {x}^{2} + - - + {( - 1)}^{n - 1} C_{n} {x}^{n}$

$\rm :\longmapsto\: {(1 - x)}^{n} =1 - C_{1}x + C_{2} {x}^{2} + - - + {( - 1)}^{n - 1} C_{n} {x}^{n}$

$\rm :\longmapsto\: {(1 - x)}^{n} - 1 = - C_{1}x + C_{2} {x}^{2} + - - + {( - 1)}^{n - 1} C_{n} {x}^{n}$

On dividing by (-1), we get

$\rm :\longmapsto\: 1 - {(1 - x)}^{n} = C_{1}x - C_{2} {x}^{2} - - - + {( - 1)}^{n - 1} C_{n}$

On dividing both sides by x, we get

$\rm :\longmapsto\: \dfrac{1 - {(1 - x)}^{n}}{x} = C_{1} - C_{2} {x}^{} - - - + {( - 1)}^{n - 2} C_{n} {x}^{n - 1}$

Now on integrating both sides between the limits x = 0 and x = 1, we get

$\rm :\longmapsto\:\displaystyle\int_0^1\dfrac{1 - {(1 - x)}^{n}}{x} = \displaystyle\int_0^1C_{1} - C_{2} {x}^{} - - - + {( - 1)}^{n - 2} C_{n} {x}^{n - 1}$

Now, Consider LHS of above integral

$\rm :\longmapsto\:\displaystyle\int_0^1\dfrac{1 - {(1 - x)}^{n}}{x}$

To integrate such integral, we use method of Substitution

So, Substitute

$\rm :\longmapsto\:1 - x = y$

$\rm :\longmapsto\:- dx = dy$

$\rm :\longmapsto\:dx = - dy$

So, above expression can be rewritten as

$\rm \:  =  \: - \displaystyle\int_{1}^0 \frac{1 - {y}^{n}}{1 - y} dy$

$\rm \:  =  \: \displaystyle\int_{0}^1 \frac{1 - {y}^{n}}{1 - y} dy$

Using Binomial theorem, we have

$\rm \:  =  \: \displaystyle\int_{0}^1 (1 + y + {y}^{2} + - - - + {y}^{n - 1}) \: dy$

$\rm \:  =  \: \bigg[y + \dfrac{ {y}^{2} }{2} + \dfrac{ {y}^{3} }{3} + - - - + \dfrac{ {y}^{n} }{n} \bigg]_{0}^1$

$\rm \:  =  \: 1 + \dfrac{1}{2} + \dfrac{1}{3} + - - - + \dfrac{1}{n}$

Now, Consider RHS of above integral

$\rm :\longmapsto \: \displaystyle\int_0^1C_{1} - C_{2} {x}^{} - - - + {( - 1)}^{n - 1} C_{n} {x}^{n - 1}$

$\rm \:  =  \: \bigg[C_{1}x - C_{2}\dfrac{ {x}^{2} }{2} + C_{3}\dfrac{ {x}^{3} }{3} + - - - + {( - 1)}^{n - 1} C_{n} \dfrac{ {x}^{n} }{n} \bigg]_{0}^1$

$\rm \:  =  \: C_{1} - \dfrac{C_{2}}{2} + \dfrac{C_{3}}{3} + - - - - + \dfrac{ {( - 1)}^{n - 1} \: C_{n}}{n}$

Hence, From above we concluded that

$\rm \: \: C_{1} - \dfrac{C_{2}}{2} + \dfrac{C_{3}}{3} + - - - - + \dfrac{ {( - 1)}^{n - 1} \: C_{n}}{n} \\ \rm \:  =  \: 1 + \dfrac{1}{2} + \dfrac{1}{3} + - - - + \dfrac{1}{n}$

$\displaystyle \sf \red{ \rm\implies \:\sum \limits^{n} _{k = 1} \binom{n}{k} \frac{( - 1) ^{k - 1} }{k} = 1 + \dfrac{1}{2} + \dfrac{1}{3} + - - - + \dfrac{1}{n} }$