Can someone please check my solution

Question

vishweshshrimali5 · Answer

Here is the question

vishweshshrimali5 · Answer

This is my answer for first part as well as an extra attempt to find the answer for a more general sequence

$$\large{S_{25} = \cfrac{1}{\log_{2} N} + \cfrac{1}{\log_{3} N} + ... + \cfrac{1}{\log_{25} N}}\tag{1}$$

$$\large{\implies S_{25} = \sum_{i=2}^{25} \cfrac{1}{\log_{i} N}}$$

$$\large{\implies S_{25} = \sum_{i=2}^{25} \log_{N} i}$$

$$\large{\implies S_{25} = \log_{N} (25!)}\tag{Ans 1}$$

$$\large{S_{n} = \log_{N} n!}\tag{Ans 2}$$

vishweshshrimali5 · Answer

Answer to second part:

$$\large{T_{25} = \cfrac{1}{\log_{2} N} - \cfrac{1}{\log_{3} N} + ... - \cfrac{1}{\log_{25} N}}$$

$$\large{\implies T_{25} = \sum_{i=2}^{25} \cfrac{(-1)^i}{\log_{i} N}}$$

$$\large{\implies T_{25} = \log_{N} [\cfrac{2 \times 4 \times 6 \times ...\times 24}{1\times 3 \times 5 \times ... \times 25}]}$$

$$\large{\implies T_{25} = \log_{N} [\cfrac{2^{12} \times 12!}{25!} \times (2^{12} \times 12!)]}$$

$$\large{\implies T_{25} = \log_{N} [\cfrac{2^{12} \times (12!)^2}{25!}]}\tag{Ans 3}$$

vishweshshrimali5 · Answer

Answer to more general version of second part:

$$\large{T_{n} = \sum_{i=1}^{n} \cfrac{(-1)^i}{\log_{i} N}}$$

$$\large{\implies T_{n} = \log_{N}[\cfrac{2^{\lfloor{\cfrac{n}{2}}\rfloor \times 2}\times (\lfloor{\cfrac{n}{2}}\rfloor !)^2}{n!}]}\tag{Ans 4}$$

Where:

$$\large{\lfloor{x}\rfloor  = \text{Floor function}}$$

vishweshshrimali5 · Answer

@ganeshie8 @ikram002p @Kainui

solomonzelman · Answer

Oh Jes..... I wish I was able to check that !

vishweshshrimali5 · Answer

Its okay :)

@SolomonZelman can you give me a latex suggestion ? My `[ ]` cannot cover the complete expression.

solomonzelman · Answer

What exactly do you want to write ?

vishweshshrimali5 · Answer

Well like this step:

$$\large{\implies T_{25} = \log_{N} [\cfrac{2^{12} \times (12!)^2}{25!}]}\tag{Ans 3}$$

See the `[ ]` ^^^

vishweshshrimali5 · Answer

I want them to cover the complete expression...

ganeshie8 · Answer

$$\large{\implies T_{25} = \log_{N} \left[\cfrac{2^{12} \times (12!)^2}{25!}\right]}\tag{Ans 3}$$
like this ?

vishweshshrimali5 · Answer

YEAH ^^^

solomonzelman · Answer

You want [ ] to be taller ?

vishweshshrimali5 · Answer

Yep

solomonzelman · Answer

ganeshie8 just gave it to you (in a good way )

vishweshshrimali5 · Answer

Yeah :)

I will try to use it more often next time :)

Thanks

vishweshshrimali5 · Answer

I have skipped maaaaaaaaaaany steps which I used to obtain the solution for general part of second part.

Actually in beginning I was trying to find out two solutions - one for odd n and one for even n

But in the end, I decided to use floor function to decide it

ganeshie8 · Answer

nice xD

vishweshshrimali5 · Answer

Thanks :)

vishweshshrimali5 · Answer

Here were my expressions for odd and even n (second part):

$$\large{T_{n} = \log_{N} \left[\cfrac{2^{n-1} \times \left(\cfrac{n-1}{2}!\right)^2}{n!}\right]}\tag{For odd n}$$

$$\large{T_{n} = \log_{N} \left[\cfrac{2^{n} \times \left(\cfrac{n}{2}!\right)^2}{n!}\right]}\tag{For even n}$$

ganeshie8 · Answer

$$\large{\implies T_{n} = \log_{N}[\cfrac{2^{\lfloor{\cfrac{n}{2}}\rfloor \color{red}{\times 2}}\times (\lfloor{\cfrac{n}{2}}\rfloor !)^2}{n!}]}\tag{Ans 4}$$

ganeshie8 · Answer

that 2 was a typo right

vishweshshrimali5 · Answer

Yeah actually that was because the the term 2^(floor(n/2)) is multiplied twice

Nope no typo

ganeshie8 · Answer

I see ! nice abuse of factorials lol

vishweshshrimali5 · Answer

Compare the general one to the special cases' answers:

$\color{blue}{\text{Originally Posted by}}$ @vishweshshrimali5
Here were my expressions for odd and even n (second part):

$$\large{T_{n} = \log_{N} \left[\cfrac{2^{n-1} \times \left(\cfrac{n-1}{2}!\right)^2}{n!}\right]}\tag{For odd n}$$

$$\large{T_{n} = \log_{N} \left[\cfrac{2^{n} \times \left(\cfrac{n}{2}!\right)^2}{n!}\right]}\tag{For even n}$$
$\color{blue}{\text{End of Quote}}$



Probably there would have been a mistake for odd n part :)

solomonzelman · Answer

Yes... I also prefer `\rm` for the words

solomonzelman · Answer

btw

1) Click and hold ALT
2) click the number code (using the numbers that are on the right of the keyboard, not the once below F1, F2, F3, etc., ) 3) release the ALT

0 2 1 5                   ×
2 4 6                      ÷
7 5 4                      ≥
7 5 5                      ≤
2 4 1       7 5 3       ±
2 4 7                      ≈
2 5 1                      √

vishweshshrimali5 · Answer

Okay

solomonzelman · Answer

just if you need.

solomonzelman · Answer

and 2 5 3 for ²

vishweshshrimali5 · Answer

Yeah I had read that somewhere but always forget that :)

solomonzelman · Answer

I have more

vishweshshrimali5 · Answer

I would have to go now.. Bye ::)

solomonzelman · Answer

`∛   ✂   ∜ `

solomonzelman · Answer

Bye ... !