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Question: If *n* is an integer greater than 1, then \[a -^{n} ⥂ C_{1}(a - 1) +^{n} ⥂ C_{2}(a - 2) - ........ ...

If n is an integer greater than 1, then

a−n⥂C1(a−1)+n⥂C2(a−2)−........+(−1)n(a−n)=a -^{n} ⥂ C_{1}(a - 1) +^{n} ⥂ C_{2}(a - 2) - ........ + ( - 1)^{n}(a - n) =

A

a

B

0

C

a2a^{2}

D

2n2^{n}

Answer

0

Explanation

Solution

We have a[C0−C1+C2......]+[C1−2C2+3C3.....]a\lbrack C_{0} - C_{1} + C_{2}......\rbrack + \lbrack C_{1} - 2C_{2} + 3C_{3}.....\rbrack

= a[C0−C1+C2......]−[−C1+2C2−3C3+......]a\lbrack C_{0} - C_{1} + C_{2}......\rbrack - \lbrack - C_{1} + 2C_{2} - 3C_{3} + ......\rbrack

We know that (1−x)n=C0−C1x+C2x2......(−1)nCnxn(1 - x)^{n} = C_{0} - C_{1}x + C_{2}x^{2}......( - 1)^{n}C_{n}x^{n};

Put x=1x = 1, 0=C0−C1+C2.......0 = C_{0} - C_{1} + C_{2}.......

Then differentiating both sides w.r.t. to x ,

we get n(1−x)n−1=0−C1+2C2x−3C3x2+.....n(1 - x)^{n - 1} = 0 - C_{1} + 2C_{2}x - 3C_{3}x^{2} + .....

Put x=1x = 1, 0=−C1+2C2−3C3+....0 = - C_{1} + 2C_{2} - 3C_{3} + .... = a[0]−[0]=0a\lbrack 0\rbrack - \lbrack 0\rbrack = 0.