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EDIT: there was an error in the previous answer . Now i think it is right. sorry for the mistake.

$2^{ rac{x}{2}} + (sqrt{2} + 1)^x =(5 + 2 sqrt{2})^{ rac{x}{2}}$

=> $(\sqrt2)^x + (\sqrt{2} + 1)^x =[(\sqrt2)^2+(\sqrt{2} + 1)^2]^{\frac{x}{2}}$

Obviously , x=2 is a solution.

Now, this is of the form, $a^{\frac{x}{2}} + b^{\frac{x}{2}} =(a + b)^{\frac{x}{2}}$

It can be proved that this does not have any other solution by using inequalities.

So x=2 is the only solution.

There is no obvious business. we r supposed to prove it . saying that it is obvious is not a proof.

suppose I want to prove that u r wrong , can i just say that u r obviously wrong?

dream is right.

in general , whenever u square an equation , u should substitute the values and check if it satisfies the equation. u need to check whether it is completely correct . even if all the terms inside the square root are positive, the answer might be still wrong.

u don't have to update it any more : here's the complete list en.wikipedia.org/wiki/Google's_hoaxes

General term,
$T_k= rac{k^2}{(k+1)!}$

$= rac{k^2-1+1}{(k+1)!}$

= rac{k-1}{(k)!} + rac{1}{(k+1)!}

= rac{1}{(k-1)!} - rac{1}{k!}+ rac{1}{(k+1)!}

Now ,
$\sum_{k=1}^\infty \frac{1}{(k-1)!} - \frac{1}{k!}+ \frac{1}{(k+1)!}$ = (e) - (e-1) +(e-2)

= e-1

Is the answer 'e'?

and should n't it be 1²/2! + 2²/3! + 3²/4! +.

Let $f(x) = x^4 + \frac{1}{x^2}$
find f '(x) and find the minima. It is obtained at $x={2^{-\frac{1}{6}}}$

The given answer is obviously wrong. the total no. of ways of getting marks ( without considering any constraints) is 51x51x101.
This is much less than 650000

u don't have to do anything until 10th april
see here

7^(7^7) and (7^7)^7 are different .

(7^7)^7 = 7^49 . The seventh root of this no. is 7^7.

But the question asked is 7^(7^7)
It's 7th root is 7^[(7^7)/7] which is equal to 7^(7^6)

There are 6 gaps into which 10 people have to be put . each of the gaps should have atleast 1 person.

this is equal to the no . of positive solutions to the equation :
x1 + x2 + .... + x6 = 10

So answer = ^10-1C_6-1 = 126.

Note that the chairman is given so he will not change. if u put the solutions of x1 , x2 ... into the gaps , it will correspond to exactly 1 solution.

All the last years answers with marking scheme http://cbse.nic.in/curric~1/Computer%20Science-Marking%20Schemes%20Science%20Subjects-XII-2007.pdf

dont say that u copied from here http://www.artofproblemsolving.com/Forum/viewtopic.php?t=190940&start=360

Follow my answer , that's all

I am getting ,

$\frac{2}{3} [ 2^{2n+1} \frac{n-1}{n+2} +1]$

\frac{2}{3} [ 2^{2n+1} \frac{n-1}{n+2} +1]

The expression can be written as

$\sum_{k=1}^{n} [4^k - \frac{4^{k+1}}{k+2} + \frac{4^{k}}{k+1} ]$

\sum_{k=1}^{n} [4^k - \frac{4^{k+1}}{k+2} + \frac{4^{k}}{k+1} ]

Now use telescopic cancellation and sum of GP