Question

If $a$ an $d$ are two complex numbers, then find the sum to $(n+1)$ terms of the series
$a{C}_0-(a+d)C_1+(a+2d)C_2-(a+3d)C_3+....,$ where $C_r={{}^{n}C}_r$

Answer 1

$(1+x{)}^n={(}^n{C}_0{x}^0{+}^n{C}_1{x}^1{+}^n{C}_2{x}^2.........{.}^n{C}_n{x}^n)\left(a\right)$

So When $x=-1{}^{}{}^n{C}_0{-}^n{C}_1{+}^n{C}_2-...........{.}^n{C}_n=0......\left(1\right)$

Differentiate Equation with respect to x once we get$n{(1+x)}^{n-1}{=}^n{C}_1{x}^0+2^n{C}_2{x}^1+..........+n^n{C}_n{x}^{n-1}$ So When $x=-1$

${}^n{C}_0{1}^0-2^n{C}_2{1}^1+..........+n^n{C}_n{1}^{n-1}=0......\left(2\right)$

Multiply Equation (1) by a and equation (2) by d and Subtract them we get$a{(}^n{C}_0{-}^n{C}_1{+}^n{C}_2-...........{.}^n{C}_n)-d{(}^n{C}_0{1}^0-2^n{C}_2{1}^1+..........+n^n{C}_n{1}^{n-1})=0a{C}_0-(a+d)C_1+(a+2d)C_2..............=0$

This is the desired Result.