First slide

Binomial theorem for positive integral Index

Question

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

Moderate

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

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