xC0−x+1C1+x+2C2−......+(−1) nx+nCn=_______ where Cr stands for nCr.

Question

xC0​​−x+1C1​​+x+2C2​​−......+(−1) nx+nCn​​=_______ where Cr​ stands for  n​Cr​.

Filo · Accepted Answer

xc0​​−x+1c1​​+x+2c2​​−............+(−1) nx+ncn​​ (1−y) n=c0​−c1​y+c2​y2−......+cn​ynyx−1(1−y) n=c0​yx−1−c1​yx+c2​yx+1−......∫01​yx−1(1−y) ndy=∫01​c0​yx−1dy−∫01​c1​yxdy+..........⇒∫01​yx−1(1−y) ndy=xc0​​−x+1c1​​+.......+(−1) nx+ncn​​ We have to calculate the the value of integration in order to find the required value of RHS. ∫01​1yx−1​ (1−y) n dy ...... (using by parts)=[xyx​(1−y) n] 0​+n∫01​xyx​(1−y) n−1dy = 0+xn​∫01​yx (1−y) n−1dyAgain by using by parts:- =xn​⋅[x+1yx+1​⋅(1−y) n−1] 0​+∫01​(n−1)(1−y) n−2x+1yx+1​  =xn​⋅[0+x+1n−1​∫01​yx+1(1−y) n−2dy] =x(x+1) n(n−1) ​∫01​yx+1(1−y) n−2dy This integration by parts will go untill  (1−y) n−2becomes(1−y) 0i.e. 1 and then the integration will stop.So, final integration will be∫01​yx+n−1dy=x+nyx+n​=x+n1​ Thus, ∫01​yx−1(1−y)n=x(x+1)(x+2)....(x+n) n(n−1)(n−2).....1​=x(x+1)...(x+n) n!​Hence, the correct option is C.

Question Text	$\frac{C _{0}}{x} - \frac{C _{1}}{x + 1} + \frac{C _{2}}{x + 2} - ...... + (- 1)^{n} \frac{C _{n}}{x + n} =$ _______ where $C_{r}$ stands for $_{n} C_{r}$ .
Answer Type	Text solution:1
Upvotes	150

$\frac{C _{0}}{x} - \frac{C _{1}}{x + 1} + \frac{C _{2}}{x + 2} - ...... + (- 1)^{n} \frac{C _{n}}{x + n} =$ _______ where $C_{r}$ stands for $_{n} C_{r}$ .

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xC0​​−x+1C1​​+x+2C2​​−......+(−1) nx+nCn​​=_______ where Cr​ stands for n​Cr​.

Text solutionVerified

Practice questions from similar books

$\frac{C _{0}}{x} - \frac{C _{1}}{x + 1} + \frac{C _{2}}{x + 2} - ...... + (- 1)^{n} \frac{C _{n}}{x + n} =$ _______ where $C_{r}$ stands for $_{n} C_{r}$ .