- n -1 n 1-log 2 n a -A- nn-1-2log 2 aB- None of theseC- nn-1-2log a 2D- n -12 n 2-4log 2 a

The correct option is C n(n+1)/2log_a 2∑_n=1^n1/log_2^n (a) = ∑_n=1^n log_a 2^n = ∑_n=1^nn log_a 2. = log_a 2. ∑_n=1^nn =log_a 2.n(n+1)/2 =n(n+1)/2log_a 2 ...

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Byju's Answer

Standard XI

Mathematics

Fundamental Laws of Logarithms

∑ n =1 n 1/lo...

Question

$\sum_{n = 1}^{n} \frac{1}{l o g_{2^{n}} (a)} =$

\frac{n (n + 1)}{2} l o g_{a} 2

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\frac{n (n + 1)}{2} l o g_{2} a

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\frac{(n + 1)^{2} n^{2}}{4} l o g_{2} a

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N o n e o f t h e s e

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Solution

The correct option is A $\frac{n (n + 1)}{2} l o g_{a} 2$
$\sum_{n = 1}^{n} \frac{1}{l o g_{2^{n}} (a)}$
$= \sum_{n = 1}^{n} l o g_{a} 2^{n}$
$= \sum_{n = 1}^{n} n l o g_{a} 2.$
$= l o g_{a} 2. \sum_{n = 1}^{n} n$
$= l o g_{a} 2. \frac{n (n + 1)}{2}$
$= \frac{n (n + 1)}{2} l o g_{a} 2$

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Similar questions

\sum_{n = 1}^{n} \frac{1}{l o g_{2^{n}} (a)} =

\sum_{n = 1}^{n} \frac{1}{l o g_{2^{n}} (a)} =

If $(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . . . . . . + C_{n} x^{2}$ , then

$C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + C_{3}^{2} + . . . . . . . . . . + C_{n}^{2}$ =

lim n \to \infty ((\frac{n}{n})^{n} + (\frac{n - 1}{n})^{n} + . . + ({\frac{1}{n}}^{n}))

equals

$n^{n} - n (n - 1)^{n} + \frac{n (n - 1)}{2!} (n - 2)^{n} - . . . \infty = \frac{3}{a} \times n!$ .

Find $a$

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∑nn=11log2n (a)=

The correct option is A n(n+1)2loga 2∑nn=11log2n (a) =∑nn=1loga 2n =∑nn=1nloga 2. =loga 2.∑nn=1n =loga 2.n(n+1)2 =n(n+1)2loga2

$\sum_{n = 1}^{n} \frac{1}{l o g_{2^{n}} (a)} =$

The correct option is A $\frac{n (n + 1)}{2} l o g_{a} 2$
$\sum_{n = 1}^{n} \frac{1}{l o g_{2^{n}} (a)}$
$= \sum_{n = 1}^{n} l o g_{a} 2^{n}$
$= \sum_{n = 1}^{n} n l o g_{a} 2.$
$= l o g_{a} 2. \sum_{n = 1}^{n} n$
$= l o g_{a} 2. \frac{n (n + 1)}{2}$
$= \frac{n (n + 1)}{2} l o g_{a} 2$