For a∈R (the set of all real numbers), a=−1),(lim)n→∞((n+1)a−1[(na+1)+(na+2)+……(na+n)]1a+2a++na=60.1Then a=(a)5 (b) 7 (c) 2−15 (d) 2−17

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For {a}{R} (the set of all real numbers), {a}-{1}{)},{()}_{{{n

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For $a \in R$ (the set of all real numbers), $a \neq = - 1),$ $(lim)_{n \to \infty} (\frac{1 ^{a} + 2 ^{a} + + n ^{a}}{( n + 1 ) ^{a - 1} [ ( n a + 1 ) + ( n a + 2 ) + \dots\dots ( n a + n ) ]} = \frac{1}{60 .}$ Then $a =$ (a) $5$ (b) 7 (c) $\frac{- 15}{2}$ (d) $\frac{- 17}{2}$

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Harper Discussed

For

a \in R

(the set of all real numbers),

a \neq = - 1),

(lim)_{n \to \infty} (\frac{1 ^{a} + 2 ^{a} + + n ^{a}}{( n + 1 ) ^{a - 1} [ ( n a + 1 ) + ( n a + 2 ) + \dots\dots ( n a + n ) ]} = \frac{1}{60 .}

Then

a =

(a)

5

(b) 7 (c)

\frac{- 15}{2}

(d)

\frac{- 17}{2}

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Question Text	For $a \in R$ (the set of all real numbers), $a \neq = - 1),$ $(lim)_{n \to \infty} (\frac{1 ^{a} + 2 ^{a} + + n ^{a}}{( n + 1 ) ^{a - 1} [ ( n a + 1 ) + ( n a + 2 ) + \dots\dots ( n a + n ) ]} = \frac{1}{60 .}$ Then $a =$ (a) $5$ (b) 7 (c) $\frac{- 15}{2}$ (d) $\frac{- 17}{2}$
Answer Type	Video solution: 1
Upvotes	150