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

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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

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We can write,$$aC0$$−(a+d)C1+(a+2d)C2$$−… upto $$(n+1) terms $$=aC0$$−$$C1+C2$$−…$$+d$$−$$C1+2C2$$−$$3C3+$$…We know,   (1$$−$$x)n=C0$$−$$C1x+C2x2$$−…$$+($$−$$1)nCnxn$$…...$$(ii) On differentiating Eq. (ii) w.r.t. x, we get−$$n(1$$−$$x)n$$−$$1=$$−$$C1+2C2x$$−…$$+($$−$$1)nCnnxn$$−$$1$$…...(iii) On putting x $$=$$ $$1$$ in Eqs. (ii) and (iii), we $$getC0$$−$$C1+C2$$−…$$+($$−$$1)nCn=0-----(iv)and$$ −$$C1+2C2$$−…$$+($$−$$1)nnCn=0----(v)from$$ eq.(i)aC0$$−$$(a+d)C1+(a+2d)C2$$−…upto $$(n+1) $$term=a$$⋅$$0+d$$⋅$$0=0$$ [fromEqs. (iv)and(v)]

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

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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

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