If a and d are two complex numbers, then the sum to (n+1) terms of the follow

Question -

If a and d are two complex numbers, then the sum

to (n+1) terms of the following series

a $C_{0}$ - (a + d) $C_{1}$ + (a + 2d) $C_{2}$ - ........... is

Options:

A .

\frac{a}{2^{n}}

B . na

C . 0

D . None of these

Answer: Option C
:
C

We can write

a $C_{0}$ - (a + d) $C_{1}$ + (a + 2d) $C_{2}$ - ........ upto (n+1) terms

$= a (C_{0} - C_{1} + C_{2} - . . . . . . . .) + d (- C_{1} + 2 C_{2} - 3 C_{3} + . . . . . .)$ .............(i)

Again, $(1 - x)^{n} = C_{0} - C_{1} x + C_{2} x^{2} - . . . . . . . . . + (- 1)^{n} C_{n} x^{n}$ ..........(ii)

Differentiating with respect to x,

$- n (1 - x)^{n - 1}$ = - $C_{1}$ + 2 $C_{2}$ x - .......... + $(- 1)^{n} C_{n} n x^{n - 1}$ ......................(iii)

Putting x = 1 in (ii) and (iii), we get

$C_{0}$ - $C_{1}$ + $C_{2}$ - ........ + (-1) $^{n} C_{n}$ = 0

and - $C_{1}$ + 2 $C_{2}$ - ........+ $(- 1)^{n} n . C_{n}$ = 0

Thus the required sum to (n+1) terms, by (i)

= a.0 + d.0 = 0.

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