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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+2 d) C_2-\ldots \ldots is

Option: 1

\frac{a}{2^n}


Option: 2

na


Option: 3

0


Option: 4

None of these


Answers (1)

best_answer

We can write a C_0-(a+d) C_1+(a+2 d) C_2-\ldots . \text { upto }(n+1) terms =a\left(C_0-C_1+C_2-\ldots\right)+d\left(-C_1+2 C_2-3 C_3+\ldots\right)

Again, (1-x)^n=C_0-C_1 x+C_2 x^2-\ldots .+(-1)^n C_n x^n

Differentiating with respect to x,

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

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

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

and -C_1+2 C_2-\ldots .+(-1)^n n \cdot C_n=0

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

=a.0 + d.0 = 0.

 

Posted by

manish painkra

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