If a and d are two complex numbers,then the s | vedclass

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(C) The given series is $S = a{C_0} - (a + d){C_1} + (a + 2d){C_2} - \dots + (-1)^n(a + nd){C_n}$.We can rewrite this as:$S = a({C_0} - {C_1} + {C_2}

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(C) The given series is $S = a{C_0} - (a + d){C_1} + (a + 2d){C_2} - \dots + (-1)^n(a + nd){C_n}$.We can rewrite this as:$S = a({C_0} - {C_1} + {C_2} - \dots + (-1)^n{C_n}) + d(0 \cdot {C_0} - 1 \cdot {C_1} + 2 \cdot {C_2} - \dots + (-1)^n n \cdot {C_n})$.Using the binomial expansion $(1 - x)^n = {C_0} - {C_1}x + {C_2}x^2 - \dots + (-1)^n{C_n}x^n$,at $x = 1$,we get:${C_0} - {C_1} + {C_2} - \dots + (-1)^n{C_n} = (1 - 1)^n = 0$.Differentiating with respect to $x$:$-n(1 - x)^{n-1} = -{C_1} + 2{C_2}x - 3{C_3}x^2 + \dots + (-1)^n n{C_n}x^{n-1}$.At $x = 1$,we get:$-{C_1} + 2{C_2} - 3{C_3} + \dots + (-1)^n n{C_n} = -n(1 - 1)^{n-1} = 0$.Substituting these values into the expression for $S$:$S = a(0) + d(0) = 0$.

If $a$ and $d$ are two complex numbers,then the sum to $(n + 1)$ terms of the following series $a{C_0} - (a + d){C_1} + (a + 2d){C_2} - \dots$ is

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