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$$\frac{1}{\log_24}+\frac{1}{\log_44} + \frac{1}{\log_84}.....\frac{1}{\log_{2^n}4}$$

MY SOLUTION

We can write it as

$$\frac{\log2}{\log4} + \frac{\log4}{\log4} + ....\frac{\log 2^n}{\log4}$$ $$=\frac{1}{\log4}\left[\log 2 + \log 4....\log 2^n\right]$$ $$=\frac{1}{\log 4}\left [\log(2.4.8....2^n)\right]$$ $$\frac{1}{\log 4}[log(2^n.2^{\frac{(n)(n+1)}{2}} )]$$

$$\frac{2n+n^2+n}{4}$$

But the answer is $\frac{n^2+n}{4}$

Aditya
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3 Answers3

1

You are almost right, we have that

$$\dots=\frac{1}{\log 4}\left [\log(2\cdot 4\cdot 8\cdot \ldots \cdot 2^n)\right]=\frac{1}{\log 4}\left [\log(2\cdot 2^2\cdot 2^3\cdot \ldots \cdot 2^n)\right]=$$

$$=\frac{1}{\log 4}\left [\log(2^{1+2+3+\dots+n} )\right]=\frac{1}{\log 4}\left [\log(2^{\frac{n(n+1)}2})\right]=\frac{n(n+1)}2\frac{\log 2}{\log 4}=\frac{n(n+1)}4$$

indeed $\frac{\log 2}{\log 4}=\frac12 \frac{2\log 2}{\log 4}=\frac12 \frac{\log 4}{\log 4}=\frac12$.

user
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  • Isn’t product of n terms of a GP $a^n.r^{\frac{(n)(n+1)}{2}}$ – Aditya Oct 21 '19 at 15:20
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    @Aditya We have $$2\cdot 4\cdot 8\cdot \ldots \cdot 2^n=2\cdot 2^2\cdot 2^3\cdot \ldots \cdot 2^n=2^{1+2+3+\dots+n}$$ – user Oct 21 '19 at 15:22
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    It may sound dumb, but isn’t it still a geometric progression? So the product formula should still work p – Aditya Oct 21 '19 at 15:24
  • I am 90% convinced I am wrong, but could you just explain that part? – Aditya Oct 21 '19 at 15:25
  • @Aditya I've added some intermediate steps. As already explained we have that the expression in the $log$ is equal to $2^{1+2+3+...+n}$ and from here we can find the given result. – user Oct 21 '19 at 15:28
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    @Aditya Let consider $2\cdot 2^2\cdot 2^3=2\cdot 4 \cdot 8=64$ which is indeed equal to $2^{\frac{3\cdot 4}2}=2^6=64$ – user Oct 21 '19 at 15:33
  • The formula you are applying is perfect but your use of the formula is incorrect. Notice that r=2 and the series starts with 2. Hence the first term (a) is 1. So the product is 1.2.4.8.... therefore while applying the formula substitute a=1 – srswat Oct 21 '19 at 15:37
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    Or you could take a=2 but then you must take n-1 as the number of terms. – srswat Oct 21 '19 at 15:41
  • Yeah that makes sense – Aditya Oct 21 '19 at 16:22
1

The series can be written by reversing the bases,

$$\log_{4} (2)+\log_{4} (4) +\log_4 (8)+\cdots$$

$$\log_{4}(2.4.8...)=\log_{4}(2^{\frac{n (n+1)}{2}}) $$

Which is $$\frac {n^2+n}{4} $$

There is an error in the product $2\cdot4\cdot8\cdots$ you took in your solution .

It can be written as $$2\cdot2^2\cdot2^3...=2^{1+2+3+...} $$ Which is $$2^{\frac{n (n+1)}{2}}$$

sai-kartik
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srswat
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0

$\dfrac{1}{2\log 2}[\log 2(1+2+3+....n)]=$

$\dfrac{n(n+1)}{2\cdot 2}$;

Used:

1)$\log 4= \log 2^2=2\log/2$;

2)$\log 2+\log 2^2+.......\log 2^n=$

$\log 2 +2\log 2 +....+n\log 2=$

$\log 2(1+2+3+.....n)=$

$\log 2\dfrac{n(n+1)}{2}$.

Peter Szilas
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