Given ${^{8} C}_{1} x {(1 - x)}^{7} + 2. {^{8} C}_{2} x^{2} {(1 - x)}^{6} + 3. {^{8} C}_{2} x^{3} {(1 - x)}^{5} + . . . + 8. x^{8} = a x + b$ Find $a + b$

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Solution

Let
$S =^{8} C_{1} x {(1 - x)}^{7} + 2^{8} C_{2} x^{2} {(1 - x)}^{6} + . . . . + 8 x^{8} = a x + b$
$S = x . \frac{d}{d x} {(x + (1 + x))}^{8} = x . \frac{d}{d x} {(1)}^{8} = 0 = a x + b$
So, $a = 0, b = 0$
Thus, $a + b = 0$

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