Question

challenge For all Aryabhatta's and Moderators .5 Stars , 2 Aryabhatta and 2 Moderators are watching my this question but don't Gave the answer .

Ans : 7 , -17/2 .

Don't Spam . Did it on your own intelligence
. It is 7 not -7.

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Answer 1

GIVEN :–

• A limit –

$\\ \implies \bf \lim_{n \to \infty } \dfrac{ {1}^{a} + {2}^{a} +. . . . . . . + {n}^{a}}{(n + 1)^{a - 1}[(na + 1) + (na + 2) +. . . . . . . + (na + n)]} = \dfrac{1}{60}$

TO FIND :–

• Value of 'a' = ?

SOLUTION :–

$\\ \implies \bf \lim_{n \to \infty } \dfrac{ {1}^{a} + {2}^{a} +. . . . . . . + {n}^{a}}{(n + 1)^{a - 1}[(na + 1) + (na + 2) +. . . . . . . + (na + n)]} = \dfrac{1}{60}$

$\\ \implies \bf \lim_{n \to \infty } \dfrac{ {1}^{a} + {2}^{a} + . . . . . . . + {n}^{a}}{(n + 1)^{a - 1}[(na + na +. . . . . . . +n - times) + (1 + 2 +. . . . . . . + n)]} = \dfrac{1}{60}$

$\\ \implies \bf \lim_{n \to \infty } \dfrac{ {1}^{a} + {2}^{a} +. . . . . . . + {n}^{a}}{(n + 1)^{a - 1} \left[n(na) + \dfrac{n(n + 1)}{2} \right]} = \dfrac{1}{60}$

$\\ \implies \bf \lim_{n \to \infty } \dfrac{ {1}^{a} + {2}^{a} +. . . . . . . + {n}^{a}}{(n + 1)^{a - 1} \left[ {n}^{2}a+ \dfrac{n(n + 1)}{2} \right]} = \dfrac{1}{60}$

• We know that –

$\\ \longrightarrow \bf 1 + 2 +. . . . . . . + n = \dfrac{n(n + 1)}{2} = \dfrac{{n}^{2} }{2} + \dfrac{n}{2}$

$\\ \longrightarrow \bf 1^{2} + 2^{2} + . . . . . . . + n^{2} = \dfrac{n(n + 1)(2n + 1)}{6} = \dfrac{{n}^{3} }{3} + \dfrac{ {n}^{2} }{c} +. . . . . . . \: \: \: \: \: \: \: \: \: [c \to constant]$

$\\ \longrightarrow \bf 1^{3} + 2^{3} + . . . . . . . + n^{3} = \left[\dfrac{n(n + 1)}{2} \right]^{2} = \dfrac{{n}^{4} }{4} + \dfrac{ {n}^{3} }{c} +. . . . . . . \: \: \: \: \: \: \: \: \: [c \to constant]$

• Similarly –

$\\ \longrightarrow \bf 1^{a} + 2^{a} + . . . . . . . + n^{a} = \dfrac{{n}^{a + 1} }{a + 1} + \dfrac{ {n}^{a} }{c} +. . . . . . . \: \: \: \: \: \: \: \: \: [c \to constant]$

• So that –

$\\ \implies \bf \lim_{n \to \infty } \dfrac{ \dfrac{{n}^{a + 1} }{a + 1} + \dfrac{ {n}^{a} }{c} +. . . . . . . }{ {(n)}^{a - 1} \left(1 + \dfrac{1}{n} \right)^{a - 1} \left[ {n}^{2}a+ \dfrac{n(n + 1)}{2} \right]} = \dfrac{1}{60}$

$\\ \implies \bf \lim_{n \to \infty } \dfrac{(n)^{a + 1} \left[\dfrac{1}{a + 1} + \dfrac{1}{nc} +. . . . . . . \right]}{ {(n)}^{a - 1} \left(1 + \dfrac{1}{n} \right)^{a - 1} \left[ {n}^{2}a+ \dfrac{n(n + 1)}{2} \right]} = \dfrac{1}{60}$

$\\ \implies \bf \lim_{n \to \infty } \dfrac{(n)^{a + 1} \left[\dfrac{1}{a + 1} + \dfrac{1}{nc} +. . . . . . . \right]}{ {(n)}^{a - 1} \left(1 + \dfrac{1}{n} \right)^{a - 1} ( {n}^{2} )\left[a+ \dfrac{(1 +1/n)}{2} \right]} = \dfrac{1}{60}$

$\\ \implies \bf \lim_{n \to \infty } \dfrac{(n)^{a + 1} \left[\dfrac{1}{a + 1} + \dfrac{1}{nc} +. . . . . . . \right]}{ {(n)}^{a + 1} \left(1 + \dfrac{1}{n} \right)^{a - 1} \left[a+ \dfrac{(1 +1/n)}{2} \right]} = \dfrac{1}{60}$

$\\ \implies \bf \lim_{n \to \infty } \dfrac{\left[\dfrac{1}{a + 1} + \dfrac{1}{nc} + . . . . . . . \right]}{\left(1 + \dfrac{1}{n} \right)^{a - 1} \left[a+ \dfrac{(1 +1/n)}{2} \right]} = \dfrac{1}{60}$

• Now apply limit –

$\\ \implies \bf \dfrac{\left[\dfrac{1}{a + 1} + \dfrac{1}{c( \infty )} +. . . . . . . \right]}{\left(1 + \dfrac{1}{ \infty } \right)^{a - 1} \left[a+ \dfrac{(1 +1/ \infty )}{2} \right]} = \dfrac{1}{60}$

$\\ \implies \bf \dfrac{\left[\dfrac{1}{a + 1} + 0 +. . . . . . . \right]}{\left(1 +0 \right)^{a - 1} \left[a+ \dfrac{(1 +0)}{2} \right]} = \dfrac{1}{60}$

$\\ \implies \bf \dfrac{\left[\dfrac{1}{a + 1}\right]}{\left(1\right)^{a - 1} \left[a+ \dfrac{1}{2} \right]} = \dfrac{1}{60}$

$\\ \implies \bf \dfrac{\left[\dfrac{1}{a + 1}\right]}{\left[a+ \dfrac{1}{2} \right]} = \dfrac{1}{60}$

$\\ \implies \bf60\left[\dfrac{1}{a + 1}\right]= \left[a+ \dfrac{1}{2} \right]$

$\\ \implies \bf\dfrac{60}{a + 1}= a+ \dfrac{1}{2}$

$\\ \implies \bf\dfrac{60}{a + 1}= \dfrac{2a + 1}{2}$

$\\ \implies \bf60 \times 2 = (2a + 1)(a + 1)$

$\\ \implies \bf 2 {a}^{2} + 2a + a + 1= 120$

$\\ \implies \bf 2 {a}^{2} +3a+ 1= 120$

$\\ \implies \bf 2 {a}^{2} +3a - 119= 0$

$\\ \implies \bf a = \frac{ - 3 \pm \sqrt{ {(3)}^{2} - 4(2)( - 119)} }{2 \times 2}$

$\\ \implies \bf a = \dfrac{ - 3 \pm \sqrt{9 + 952} }{4}$

$\\ \implies \bf a = \dfrac{ - 3 \pm \sqrt{961} }{4}$

$\\ \implies \bf a = \dfrac{ - 3 \pm 31}{4}$

$\\ \implies \bf a = \dfrac{ - 3 + 31}{4} ,\dfrac{ - 3 - 31}{4}$

$\\ \implies \bf a = \dfrac{28}{4} ,\dfrac{ - 34}{4}$

$\\ \implies\large{\boxed{\bf a =7, - \dfrac{17}{2}}}$

• Hence , The value of a = 7 ,-17/2.