ANSWER-

Hallo aditya.

Example: If n = 1 we have:

\(3^{2n+2} - 8n -9 = 3^{2\cdot 1+2} - 8\cdot 1 -9 = 3^4-8-9 = 81-17 = 64 ={\color{red}2^6}\)

\(2^6 \text{ can be devide by } 2^m \text{, if }m=6.\\ So~the~mimimum~ m~ is ~ 6.\\ \text{The maximum m must be equal or greater than the minimum m }.\\ \text{So m is }6\text{ or greater}.\)

m is not a constant.

cool :)

Hi Aditya, your answer looks really interesting so I will take a look at your working :)

n belongs to natural no's, then the highest integer m such that 2^m divides 3^2n+2-8n-9 is

=9(9^n)-8n-9

9(8+1)^n  -8n - 9

9(nc08n +nc18n-1 + ...........+ncn80) -8n -9

9(8x+1) -8n-9  (let the remaining terms be expressed as x(ie.nc08n-1,nc18n-2 etc.)

72x+9-8n-9

72x-8n

8(9x-n) divisible by 2m

2m =8

m=3

\(3^{2n+2}-8n-9\\ =9*3^{2n}-8n-9\\ =9*9^n-8n-9\\ =9*(8+1)^n-8n-9\\ =9\left[\begin{pmatrix}n\\0\end{pmatrix}8^0+\begin{pmatrix}n\\1\end{pmatrix}8^1+.......\begin{pmatrix}n\\{n-1}\end{pmatrix}8^{n-1}+\begin{pmatrix}n\\n\end{pmatrix}8^n\right]-8^n-9\\ =9\left[1+n*8^1+.......\begin{pmatrix}n\\{n-1}\end{pmatrix}8^{n-1}+8^n\right]-8^n-9\\ =9\left[8n+.......\begin{pmatrix}n\\{n-1}\end{pmatrix}8^{n-1}+8^n\right]-8^n\\\)

I think that is correct but I do not think it simplifies to what you have there.......

In response to aditya below.

You want me to factor 8 out, ok I can do that.

\(=9*8\left[n+\begin{pmatrix}n\\2\end{pmatrix} 8 +........ \begin{pmatrix}n\\n-1\end{pmatrix} 8^{n-2}+8^{n-1} \right]-8 ^n\\ =2^3*9\left[n+\begin{pmatrix}n\\2\end{pmatrix} 8 +........ \begin{pmatrix}n\\n-1\end{pmatrix} 8^{n-2}+8^{n-1} \right]-2^{3n}\\ =2^3*9\left[n+\begin{pmatrix}n\\2\end{pmatrix} 8 +........ \begin{pmatrix}n\\n-1\end{pmatrix} 8^{n-2}+8^{n-1} \right]-2^3*2^{3n-3}\\ =2^3*\left[9\left[n+\begin{pmatrix}n\\2\end{pmatrix} 8 +........ \begin{pmatrix}n\\n-1\end{pmatrix} 8^{n-2}+8^{n-1} \right]-2^{3n-3}\right]\)

So if n>1 and a natural number then this expression is divisable by 8.  

Note: if n=1 I don't think that this expression IS divisable by 8. because 8^(-1) is a fraction.

Heureka's simple substitution showed this to be incorrect so there must be something wrong  

answer to heureka's comment=

8 also divides 64

you may argue that 4, 2, 1 also divides 64

but i have already proved it above

answer to melody's comment=

when we take 8 common from  (nc₀8^n +nc₁8^n-1 + ...........+nc_n8^0) we get (nc₀^8n-1,nc₁8^n-2....)

for eg-  8 common from (1(nc₀8^n) we get= 8(8^n-1)

                                        nc₁8^n-1=8(nc₁8^n-2) etc

remaining same

what do you think

please reply back

exams are approaching

and binomial theorem carries 90 marks

At Aditya's request, I have continued my analysis on my post #3.        BUT it is not correct.

There is an error somewhere. ://

can you please explain where i went wrong in my recent explanation

i had a look at heureka's comment and replied to it

please tell

in the answer book given by IIT(indian institute of technology) which is without explanation, the answer is given as 3

so i thought i got ir right

please elaborate where i went wrong

Aditya, you are not asking the right person.

I looked at what you did and my answer was proven wrong by Heurekas simple substitution.

You have not answered in full so it is impossible to say what you did wrong but I think you have ignored a whole heap of the terms :/