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Question

Find  the coefficient of $$x^{50}$$ in the expression 
$$(1 \, + \, x )^{1000} \, + \, 2x (1 \, + \, x )^{999} \, + \, 3x^2 (1 \, + \, x ) ^{998}\, +  1000x ^{1000}$$


Solution

Let S = $$(1 \, + \, x)^{1000} \, + \, 2x \, (1 \, + \, x)^{999} \, + \, 3x^2 \, (1 \, + \, x)^{998} \, + \, ... \, + \,$$
$$1000 \, x^{999} \, (1 \, + \, x) \, + \, 1001 \, x^{1000}$$
Above is A.G.S. of common ratio r = $$\dfrac{x}{1 \, + \, x}$$
$$\therefore \, \, \, \, [x / (1 \, + \, x)] \, S \, = \, x \, (1 \, + \, x)^{999} \, + \, 2x^2 \, (1 \, + \, x)^{998} \, + $$
$$..... \, + \, 1000 \, \cdot \, x^{1000} \, + \, \dfrac{1001 \, x^{1001}}{1 \, + \, x}$$
Subtracting, 
$$\left(1 \, - \, \dfrac{x}{1 \, + \, x}\right)S \, = \, (1 \, + \, x)^{1000} \, + \, x \, (1 \, + \, x)^{999} $$
$$\, + \, x^2 \, (1 \, + \, x)^{998} \, + \, .... \, + \, x^{1000} \, - \, \dfrac{1001 \, x^{1001}}{1 \, + \, x}$$
or    S = $$(1 \, + \, x)^{1001} \, + \, x(1 \, + \, x)^{1000} \, + \, x^2 \, (1 \, + \, x)^{999} \, + $$
$$...... \, + \, x^{1000} \, (1 \, + \, x) \, - \, 1001 \, x^{1001}$$
$$= \, \dfrac{(1 \, + \, x)^{1001}[1 \, - \ (x\, (1 \, + \, x))^{10001}]}{1 \, - \, x} \, - \, 1001 \, x^{1001}$$
Sum G.P.
$$(1 \, + \, x)^{1002} \, [1 \, - \, (x / (1 \, + \, x))^{1001}] \, - \, 1001 \, x^{1001}$$
$$= \, (1 \, + \, x)^{1002} \, - \, x^{1001} \, (1 \, + \, x) \, - \, 1001 \, x^{1001}$$
$$= \, (1 \, + \, x)^{1002} \, - \, x^{1002} \, - \, 1002 \, x^{1001}$$            ....(1)
Now the coefficients of $$x^{50}$$ on the  R.H.S. of (1)
$$= \, ^{1002}C_{50}$$

Mathematics

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