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Explanation:

S = 2x + 5x2 + 9x3 + 14x4 + 20x5 + ...                              …(i)

Multiplying both sides by x, we get

xS = 2x2 + 5x3 + 9x4 + 14x5 + 20x6 + ...                             ...(ii)

Subtracting (ii) from (i),

(1 – x)S = 2x + 3x2 + 4x3 + 5x4 + 6x5 + ...                         ...(iii)

Again multiplying both the sides by x, we get

x(1 – x)S = 2x2 + 3x3 + 4x4 + 5x5 + 6x6 + ...                      ...(iv)

Subtracting (iv) from (iii),

(1 – x)S – x(1 – x)S = 2x + x2 + x3 + x4 + x5 + ...

(1 – x)2S = 2x + x2 + x3 + x4 + x5 +...

(1 – x)2S = x + (x + x2 + x3 + ... + ∞)

(1 - x)2S = x + x1-x ∵ Sa1-r for r < 1

∴ (1 - x)2S = x-x2+x1-x

∴ S = x(2-x)(1-x)3

Hence, option (a).

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