If the sum of the series ( 1 2 - 1 3 ) + ( 1 2 2 - 1 2 · 3 + 1 3 2 ) + ( 1 2 3 - 1 2 2 · 3 + 1 2 · 3 2 - 1 3 3 ) + ( 1 2 4 - 1 2 3 · 3 + 1 2 2 · 3 2 - 1 2 · 3 3 + 1 3 4 ) + . . . is , where and are co-prime, then + 3 is equal to ________ .

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Solution

Q.
If the sum of the series $(\frac{1}{2} - \frac{1}{3}) + (\frac{1}{2^{2}} - \frac{1}{2 \cdot 3} + \frac{1}{3^{2}}) + (\frac{1}{2^{3}} - \frac{1}{2^{2} \cdot 3} + \frac{1}{2 \cdot 3^{2}} - \frac{1}{3^{3}}) + (\frac{1}{2^{4}} - \frac{1}{2^{3} \cdot 3} + \frac{1}{2^{2} \cdot 3^{2}} - \frac{1}{2 \cdot 3^{3}} + \frac{1}{3^{4}}) + . . .$ is $\frac{α}{β},$ where $α$ and $β$ are co-prime, then $α + 3 β$ is equal to ________ . [2023]

Ans.
(7)

$Let S = (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{2^{2}} - \frac{1}{2 \cdot 3} + \frac{1}{3^{2}}) + (\frac{1}{2^{3}} - \frac{1}{2^{2} \cdot 3} + \frac{1}{2 \cdot 3^{2}} - \frac{1}{3^{3}}) + \dots$

$= (x - y) + (x^{2} - x y + y^{2}) + (x^{3} - x^{2} y + x y^{2} - y^{3}) + \dots, where x = \frac{1}{2}, y = \frac{1}{3}$

$Multiply both sides by x + y$

$(x + y) S = (x^{2} - y^{2}) + (x^{3} + y^{3}) + (x^{4} - y^{4}) + \dots$

$(x + y) S = (x^{2} + x^{3} + x^{4} + \dots) - (y^{2} + y^{4} + y^{6} + \dots) + (y^{3} + y^{5} + \dots)$

$= [\frac{\frac{1}{4}}{1 - \frac{1}{2}}] - [\frac{\frac{1}{9}}{1 - \frac{1}{9}}] + [\frac{\frac{1}{27}}{1 - \frac{1}{9}}]$ [infinite G.P.]

$= \frac{1}{2} - \frac{1}{8} + \frac{1}{24}$

$(\frac{1}{2} + \frac{1}{3}) \cdot S = \frac{5}{12}$

$S = \frac{5}{12} \times \frac{6}{5} = \frac{1}{2}$ $∴$ $α = 1, β = 2$ , so $(α + 3 β) = 1 + 3 \times 2 = 7$

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