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  • If a and d are two complex numbers, then the sum to (n + 1) terms of the series a − (a + d)<img alt="C_1" src="

If a and d are two complex numbers, then the sum to (n + 1) terms of the series aC_0 − (a + d)C_1 + (a + 2d)C_2 − (a + 3d)C_3 + … is
 

Option: 1

    a/2^n


Option: 2

na


Option: 3

0


Option: 4

none of these


Answers (1)

best_answer

We have 

\begin{aligned} & a C_0-(a+d) C_1+(a+2 d) C_2-(a+3 d) C_3+\ldots \text { up to }(n+1) \text { terms } \\ & =a\left(C_0-C_1+C_2-C_3+\ldots+(-1)^n \cdot C_n\right) +d\left(-C_1+2 C_2-3 C_3+\ldots+\right. \left.(-1)^n n C n\right) \end{aligned}We know that
(1-x)^n=C_0-C_1 x+C_2 x^2-\ldots+(-1)^n \cdot C_n x^n                  (1)

-n(1-x)^{n-1}=-C_1+2 C_2 x-\ldots+(-1)^n \cdot C_n n x^{n-1}         (2)

C_0-C_1+C_2-\ldots+(-1)^n C_n=0

and -C_1+2 C_2-\ldots+(-1)^n n C_n=0

Thus a C_0-(a+d) C_1+(a+2 d) C_2-\ldots  up to (n + 1) terms = a.0 + d.0 = 0

Posted by

Rishi

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