Find-n-C-1-1-2-n-C-2-1-3-n-C-3-1-n-1-1-n-n-C-n-

Question Number 23928 by Tinkutara last updated on 10/Nov/17

Find^{n} C_{1} - \frac{1}{2}^{n} C_{2} + \frac{1}{3}^{n} C_{3} - \dots . +

{(- 1)}^{n - 1} \frac{1}{n} \cdot^{n} C_{n} .

Commented by prakash jain last updated on 10/Nov/17

\frac{1 - {(1 - x)}^{n}}{x} = \frac{1}{x} (\underset{i = 1}{\sum^{n}} {(- 1)}^{i - 1}^{n} C_{i} x^{i})

\frac{1 - {(1 - x)}^{n}}{x} = (\underset{i = 1}{\sum^{n}} {(- 1)}^{i - 1}^{n} C_{i} x^{i - 1})

\int_{0}^{1} \frac{1 - {(1 - x)}^{n}}{x} d x = \int_{0}^{1} (\underset{i = 1}{\sum^{n}} {(- 1)}^{i - 1}^{n} C_{i} x^{i - 1}) d x

RHS = r e q u i r e d s u m

LHS = r e s u l t

Commented by prakash jain last updated on 10/Nov/17

Yes . Result is the value of integral

but it may not be the answer you

are looking for .

What is the answer ? We can use another

approach to reduce the expression

to required form .

Commented by prakash jain last updated on 11/Nov/17

\frac{1 - {(1 - x)}^{n}}{x} = 1 + (1 - x) + {(1 - x)}^{2} + . . + {(1 - x)}^{n - 1}

RHS is GP w i t h a = 1, r = 1 - x = \frac{1 (1 - {(1 - x)}^{n})}{1 - (1 - x)}

\int_{0}^{1} \frac{1 - {(1 - x)}^{n}}{x} d x

= {[x - \frac{{(1 - x)}^{2}}{2} - \frac{{(1 - x)}^{3}}{3} - \dots - \frac{{(1 - x)}^{n}}{n}]}_{0}^{1}

= 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}

Commented by Tinkutara last updated on 11/Nov/17

Thank you very much Sir!