Question

emergency!!!

Answer 1

which q

Answer 2

Last one x = y = z = 4

Answer 3

any question that u like...these r the question which i wsn't able to solve

Answer 4

First one probably some fermat's little theorem nonsense, the second one looks like part of the Riemann zeta function so you could probably just multiply two geometric series or something together for it, and 3rd one looks like it's probably a telescoping series, just my first guesses

Answer 5

@ParthKohli how did u get x=y=z=4

Answer 6

It's a symmetric situation. Apply all inequalities you please, but you'll finally get to this.

Answer 7

Simplest way to think of it (not a proof) is that \[n^2 > (n-a)(n+a)\] when a is not equal to zero, so that's the max when they're all the same.

Answer 8

first q doesnt look easy

Answer 9

I think we can split that 3rd one up using partial fractions, then it becomes a telescoping series that way.

Answer 10

Good questions by the way!

Answer 11

thanks @Empty
@ParthKohli i can't get to the result x=y=z.
can u tell how to start ..

Answer 12

I've only given a qualitative argument here so far - the whole situation you have here is symmetric in nature. x can be interchanged with y, y with z or z with x, and the situation won't change.

So if the triplet \((a, b , c)\) gives you a maximum value then so should \((b, a, c)\) or \((b, c, a)\) or \((c, a, b)\), which means that \(x = y = z \) when the given expression is maximum.

Let me think of an inequality that can be applied here. You'll see that we'll find the maximum when the equality of two expressions is considered, which is almost always when all variables are equal.

Answer 13

Ok so for the second one I'm getting:

$$\frac{1}{1-9}\frac{1}{1-4}\frac{5}{6}$$ Does that match your answer?

Answer 14

wait m doing the 1st one

Answer 15

for first one you may want to try factoring it
http://www.wolframalpha.com/input/?i=factor+5%5E5%5E%28n%2B1%29%2B5%5E5%5En%2B1

Answer 16

Yeah I think it is more clear how to solve the first one if you do that, here to make factoring easier substitute:

$$5^{5^n}=a$$

then you end up with the expression:

$$a^5+a+1$$

which somehow wolfram is able to factor out into:

$$(a^2+a+1)(a^3-a^2+1)$$

Answer 17

@ParthKohli
if we are given a question like -
x+y=4 and x belongs to positive integers. and we are supposed to find out maximum of
xy + x/y +y/x
then by symmetric approach x=y=2 but in this question x=3 and y=1 gives the maximum so the symmetric approach is violated.

Answer 18

@Empty we can't write 5^5^(n+1) as a^5 if a=5^5^n

Answer 19

OK, I didn't say that x = y gives you the maximum.

I said it gives you the maximum if it exists.

Here, x = y is most probably giving you the minimum.

Answer 20

yes so how do we know that in which case x=y gives the max

Answer 21

$$5^{5^{n+1}}=5^{5*5^n} = (5^{5^n})^5$$ @imqwerty

Answer 22

You have to check if x = y is giving you the maximum/minimum by entering another pair of values. If that pair gives you a higher value, then x = y gives you the minimum.

Answer 23

i feel #1 is still a stupid q, you're not gona learn anything frm it

Answer 24

wait sry yes we can put a^5...

Answer 25

yes it is @ganeshie8 :D

Answer 26

Y'know, qwerty, it's just like that with single variable functions. You can differentiate a function and equate it to zero, but you still have to check if that gives you the maxima or minima by other methods.

Answer 27

Ok so I also found a good way of solving the second one:

We only want odd total powers, so the way we can do that is only combine even powers with odd and odd with evens. So in order to do that, we multiply the geometric series' together:

$$\left(\sum_{i=0}^{\infty} \frac{1}{2^{2i}} \right) \left(\sum_{j=0}^{\infty} \frac{1}{3^{2j+1}} \right) +\left(\sum_{i=0}^{\infty} \frac{1}{3^{2i}} \right) \left(\sum_{j=0}^{\infty} \frac{1}{2^{2j+1}} \right) $$
factor out a 1/3 and 1/2 so you get only geometric series in terms of 1/4 and 1/9, and you're done!

Answer 28

Isn't that the standard way to do it though? How did you originally do it?

Answer 29

I factored it

Answer 30

the 1st one is solved thanks guys :)

Answer 31

\[\begin{align} x^5 + x + 1 & = (x^5-x^2)+(x^2+x+1)\\ \\ & = x^2(x^3 - 1) + (x^2 + x + 1) \\ \\ & = x^2(x-1)\color{blue}{\bf (x^2 + x + 1)} + \color{blue}{\bf (x^2 + x + 1)}\end{align}\]

http://math.stackexchange.com/questions/477295/factor-x5-x-1

Answer 32

To be fair, that solution was most likely obtained by working backwards. :P

Answer 33

that is okay for a proof

Answer 34

Yes of course, but you can't really think of that on your own.

Answer 35

that all goes as "scratch work"

:P

Answer 36

ahhh thanks @ganeshie8 that's a clever trick I think I can even remember that as adding fancy form of zero and then abusing the geometric series yet again lol

Answer 37

I'm ashamed that I have no other ideas other than telescoping for #3.

Answer 38

It wants to telescope, why force it to be anything it's probably not? :P

Answer 39

When you separate it out, the coefficients are both 1, it's truly a beautiful result

Answer 40

I'm either dumb or too much drunk to notice it but why do you think telescoping is a bad idea ?

Answer 41

Yeah you're not drunk enough ganeshie, cause this baby wants to see the moon

Answer 42

I didn't say telescoping is a bad idea. I just said that this series doesn't seem to be telescoping nicely and I have no other ideas.

Answer 43

m really confused nd annoyed nd kinda drunk ..cheers @ganeshie8
@Empty ..can u please tell me what u did with the 2nd question

Answer 44

my mind is jst nt wrkin normal today :(

Answer 45

OK, it's actually telescoping really nicely.

Answer 46

Here I made it smaller, part of the total sum, just write out all these terms (it's not tooo many, and it's good for you) \[\left(\sum_{i=0}^{3} \frac{1}{2^{2i}} \right) \left(\sum_{j=0}^{3} \frac{1}{3^{2j+1}} \right) +\left(\sum_{i=0}^{3} \frac{1}{3^{2i}} \right) \left(\sum_{j=0}^{3} \frac{1}{2^{2j+1}} \right) \]

Answer 47

no i mean hw did u split that

Answer 48

No, I _created_ this

Answer 49

so by multiplying it out you can kinda see how I came up with this too... I'll draw a picture of how I think about it

Answer 50

\[\frac{6^k}{(3^k - 2^k)(3^{k+1 } - 2^{k+1})}\]\[= \frac{3^{k+1}}{2^{k+1} - 3^{k+1}} - \frac{3^k}{3^k - 2^k}\]

Answer 51

There's a typo.

Answer 52

So this is the first half of the terms! @imqwerty