If ` (1 + x)^(n) = C_(0) + C_(1) x + C_(2) x^(2) + C_(3) x^(3) + …+ C_(n) x^(n)` , show that
` C_(1) - (C_(2))/(2) + (C_(3))/(3) - …(-1)^(n-1) (C_(n))/(n) = 1 + (1)/(2) + (1)/(3) + …+ (1)/(n)`.

Explanation

We know that ,
`(1 -x)^(n) = C_(0) - C_(1) x + C_(2)x^(2) -…+ (-1) ^(n) C_(n) x^(n)`
or ` C_(0) - (1 - x)^(n) = C_(1) x - C_(2)x^(2) + C_(3) x^(3)`
` -…+ (-1)^(n-1) C_(n)x^(n)`
` rArr 1 - (1 - x)^(n) = C_(1)x - C_(2)x^(2) + C_(3)x^(3) -...+ (-1)^(n-1) C_(n) x^(n)`
Dividing in each side by x , then
` (1-(1-x)^(n))/(x) = C_(1) - C_(2) x + C_(3) x^(2) -...+ (-1)^(n-1) C_(n) x^(n-1)`
On integrating within limits 0 to , we have
`int_(0)^(1) (1-(1-x)^(n))/(x) dx = int_(0)^(1) (C_(1) - C_(2) x + C_(3) x^(2) - ...+ (-1)^(n-1) C_(n) x^(n-1))dx `
`= [C_(1) x - (C_(2)x^(2))/(2) + (C_(3)x^(3))/(3) -...+ (-1)^(n-1) C_(n) (x^(n))/(n)]_(0)^(1)`
`int_(0)^(1) (1-(1-x)^(n))/(x)dx = C_(1) - (C_(2))/(2) + (C_(3))/(3) -...+ ((-1)^(n-1))/(n)C_(n)`
Putting 1 - x = t in integral
` rArr dx = - dt `
when ` x to 1, t to0 " and when " x to 0, t to 1`
`therefore int_(0)^(1) ((1-t)^(n))/((1 -t)) (-dt) = C_(1) - (C_(2))/(2) + (C_(3))/(3) -...+ (-1)^(n-1) (C_(n))/(n)`
`rArr int_(0)^(1) ((1-t)^(n))/((1 -t)) (-dt) = C_(1) - (C_(2))/(2) + (C_(3))/(3) -...+ (-1)^(n-1) (C_(n))/(n)`
`rArr int_(0)^(1) (1 +t+ t^(2) +...+ t^(n-1)) dt = C_(1)-(C_(2))/(2) + (C_(3))/(3)-...+ (-1)^(n-1) (C_(n))/(n)`
`rArr [t + (t^(2))/(2) + (t^(3))/(3) +...+ (t^(n))/(n)]_(0)^(1) = C_(0) - (C_(2))/(2) + (C_(3))/(3) -...+ (-1)^(n-1) (C_(n))/(n)`
`rArr 1 + 1/2+1/3+...+= =1/n= C_(1) - (C_(2))/(2) + (C_(3))/(3) -...+ (-1)^(n-1) (C_(n))/(n)`
Hence , `C_(1) - (C_(2))/(2) + (C_(3))/(3) -...+ (-1)^(n-1) (C_(n))/(n)`
` = 1 + 1/2+1/3+...+1/n`