ANSWER: OPTION (B) $S_r = \frac{6.x}{(1-x)^4}$
Let, $S_r = 3.2.1x + 4.3.2x^2 + 5.4.3x^3 + 6.5.4x^4 + …. $
$S_r.x = \ \ \ \ \ \\ \ \ \ \ \ \\ \ \ \ \ \\ \ \ \ 3.2.1x^2 + 4.3.2x^3 + 5.4.3x^4 + …. $ {multiplying both sides with $x$}
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$S_r-S_rx= 6.x+ 18.x^2 + 36.x^3 + 60.x^4 +….$ {Subtracting}
$S_r(1-x) = 6.x(1+3.x +6.x^2 +10.x^3$
$S_r= (6.x/1-x ) * (1+3.x +6.x^2 +10.x^3$
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$S_r = (\frac{6.x}{1-x}).(1+3x + 6.x^2+10.x^3+….)$ → $eqn(1)$
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Let $S1 = 1+3x + 6.x^2+10.x^3+….$
$S1.x = \ \ \ \ \ \ \ \ \ 1.x+3.x^2 +6.x^3 + \ ...$
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$S1(1-x)= 1+2.x+3.x^2+4.x^3+...$ → $eqn(2)$ {Subtracting}
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Let $S2= 1+2.x+3.x^2+4.x^3+...$
$S2.x= \ \ \ \ \ \ \ \ \ 1.x+2.x^2+3.x^3+...$
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$S2(1-x)= 1+x+x^2+x^3+… = \frac{1}{1-x}$ → $eqn(3)$ {Subtracting}
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Hence, $S2 = \frac{1}{(1-x)^2}$
$\implies S1 = \frac{1}{(1-x)^3}$
Thus, $S_r = \frac{6.x}{1-x}. \frac{1}{(1-x)^3} \implies S_r = \frac{6.x}{(1-x)^4}$