Let S = 2x + 5x2 + 9x3 + 14x4 + 20x5 ... infinity (x < 1)
The coefficient of nth term = n(n+3)2. The sum is
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 ∵ S∞ = a1-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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