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 ) + . . . . . .

Correct answer: 0 Given, 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 ) + . . . . . . Now let a = (1)/(2), b = (1)/(3) So, series will be: (a - b) + ( a ^ 2 - ab + b ^ 2 ) + ( a ^ 3 - a ^ 2 b + a b ^ 2 - b ^ 3 ) + . . . . . = (1)/(a + b)(( a ^ 2 - b ^ 2 ) + ( a ^ 3 + b ^ 3 ) + ( a ^ 4 - b ^ 4 ) + ) = ((1)/(a + b))( a ^ 2 + a ^ 3 + a ^ 4 - ( b ^ 2 - b ^ 3 + b ^ 4 .)) Now using infinite G . P formula we get, = ((1)/(a + b))( a ^ 2 + a ^ 3 + a ^ 4 - ( b ^ 2 - b ^ 3 + b ^ 4 .)) = ((1)/(a + b))( a ^ 2 1 - a - b ^ 2 1 + b ) Now putting the value of a and b we get, = (1)/((1)/(2) + (1)/(3))(((1)/(4))/(1 - (1)/(2)) - ((1)/(9))/(1 + (1)/(3))) = (6)/(5)((1)/(2) - (1)/(12)) = (6)/(5)((5)/(12)) = (1)/(2) = (α)/(β) Hence, α + 3β = 1 + 6 = 7

JEE Main 202315 AprilMorning Shift

Exam: JEE Main
Year: 2023
Date: 15 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2023 · 15 April · Morning Shift · Mathematics · Sequences and Series

If the sum of the series ((1)/(2) - (1)/(3)) + ( 1 2 ^ 2 - (1)/(2 ·...

If the sum of the series

$\left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \left(\dfrac{1}{{2}^{2}} - \dfrac{1}{2 \cdot 3} + \dfrac{1}{{3}^{2}}\right) + \left(\dfrac{1}{{2}^{3}} - \dfrac{1}{{2}^{2} \cdot 3} + \dfrac{1}{2 \cdot {3}^{2}} - \dfrac{1}{{3}^{3}}\right) + \left(\dfrac{1}{{2}^{4}} - \dfrac{1}{{2}^{3} \cdot 3} + \dfrac{1}{{2}^{2} \cdot {3}^{2}} - \dfrac{1}{2 \cdot {3}^{3}} + \dfrac{1}{{3}^{4}}\right) + . . . . . .$ is $\dfrac{\alpha}{\beta}$ , where $\alpha$ and $\beta$ are co-prime, then $\alpha + 3\beta$ is equal to ________.

Correct answer

Step-by-step solution

Given,
Series $\left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \left(\dfrac{1}{{2}^{2}} - \dfrac{1}{2 \cdot 3} + \dfrac{1}{{3}^{2}}\right) + \left(\dfrac{1}{{2}^{3}} - \dfrac{1}{{2}^{2} \cdot 3} + \dfrac{1}{2 \cdot {3}^{2}} - \dfrac{1}{{3}^{3}}\right) + \left(\dfrac{1}{{2}^{4}} - \dfrac{1}{{2}^{3} \cdot 3} + \dfrac{1}{{2}^{2} \cdot {3}^{2}} - \dfrac{1}{2 \cdot {3}^{3}} + \dfrac{1}{{3}^{4}}\right) + . . . . . .$
Now let $a = \dfrac{1}{2}, b = \dfrac{1}{3}$
So, series will be:
\left(a - b\right) + \left({a}^{2} - ab + {b}^{2}\right) + \left({a}^{3} - {a}^{2}b + a{b}^{2} - {b}^{3}\right) + . . . . . $$ = \dfrac{1}{a + b}\left(\left({a}^{2} - {b}^{2}\right) + \left({a}^{3} + {b}^{3}\right) + \left({a}^{4} - {b}^{4}\right) + \ldots\right) $$ = \left(\dfrac{1}{a + b}\right)\left({a}^{2} + {a}^{3} + {a}^{4} \ldots - \left({b}^{2} - {b}^{3} + {b}^{4} \ldots .\right)\right) $Now using infinite$ G . P $formula we get,$ = \left(\dfrac{1}{a + b}\right)\left({a}^{2} + {a}^{3} + {a}^{4} \ldots - \left({b}^{2} - {b}^{3} + {b}^{4} \ldots .\right)\right) $$ = \left(\dfrac{1}{a + b}\right)\left(\dfrac{{a}^{2}}{1 - a} - \dfrac{{b}^{2}}{1 + b}\right) $Now putting the value of$ a \text{ and } b $we get,$ = \dfrac{1}{\dfrac{1}{2} + \dfrac{1}{3}}\left(\dfrac{\dfrac{1}{4}}{1 - \dfrac{1}{2}} - \dfrac{\dfrac{1}{9}}{1 + \dfrac{1}{3}}\right) $$ = \dfrac{6}{5}\left(\dfrac{1}{2} - \dfrac{1}{12}\right) $$ = \dfrac{6}{5}\left(\dfrac{5}{12}\right) = \dfrac{1}{2} = \dfrac{\alpha}{\beta} $Hence,$ \alpha + 3\beta = 1 + 6 = 7

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