if a and d, are two complex numbers, then the sum to (n+ 1) terms of the following series aC0−(a+d)C1+(a+2d)C2−…+… is

A
$\frac{α}{2^{n}}$
B
na
C
0
D
None of these

Solution:

We can write,

$\begin{aligned} {aC}_{0} - (a + d) C_{1} + (a + 2 d) C_{2} - \dots upto (n + 1) terms \\ = a (C_{0} - C_{1} + C_{2} - \dots) + d (- C_{1} + 2 C_{2} - 3 C_{3} + \dots) \end{aligned}$

We know,

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

On differentiating Eq. (ii) w.r.t. x, we get

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

On putting x = 1 in Eqs. (ii) and (iii), we get

$C_{0} - C_{1} + C_{2} - \dots + (- 1)^{n} C_{n} = 0 - - - - - (iv)$

and $- C_{1} + 2 C_{2} - \dots + (- 1)^{n} {nC}_{n} = 0 - - - - (v)$

from eq.(i)

${aC}_{0} - (a + d) C_{1} + (a + 2 d) C_{2} - \dots$ upto (n+1) term

$= a \cdot 0 + d \cdot 0 = 0 [from Eqs . (iv) and (v)]$

if a and d, are two complex numbers, then the sum to (n+ 1) terms of the following series ${aC}_{0} - (a + d) C_{1} + (a + 2 d) C_{2} - \dots + \dots is$

Solution:

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