ParthKohli
  • ParthKohli
PARTY
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Error1603
  • Error1603
?
ParthKohli
  • ParthKohli
WE'RE PARTYING HERE. @GANESHIE8
ParthKohli
  • ParthKohli
If \((9+\sqrt{80})^n = I + f\) where \(I, n\) are integers and \(0

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ParthKohli
  • ParthKohli
If \[y = \frac{2}{5}+\frac{1\cdot 3}{2!}\left(\frac{2}{5}\right)^2 + \frac{1\cdot 3 \cdot 5}{3!}\left(\frac{2}{5}\right)^3 + \cdots\]Find the value of \(y^2 + 2y\).
ParthKohli
  • ParthKohli
Let \(n\) be a positive integer with\[f(n) + 1! + 2! + 3! + \cdots + n!\]and \(P(x), Q(x)\) be polynomials in \(x\) such that \(f(n+2) = P(n) f(n+1) + Q(n) f(n)\) for all \(n\ge 1\). Then: (A) P(x) = x+3 (B) Q(x) = -x - 2 (C) P(x) = -x - 2 (D) Q(x) = x + 3
anonymous
  • anonymous
@ParthKohli Are you on some new medication?
ParthKohli
  • ParthKohli
Yes.
anonymous
  • anonymous
I had a feeling.
anonymous
  • anonymous
@ParthKohli 2+2=? Solve for a cookie.
ParthKohli
  • ParthKohli
idk lol that's way harder than the questions I posted
ganeshie8
  • ganeshie8
If \[y = \frac{2}{5}+\frac{1\cdot 3}{2!}\left(\frac{2}{5}\right)^2 + \frac{1\cdot 3 \cdot 5}{3!}\left(\frac{2}{5}\right)^3 + \cdots\]Find the value of \(y^2 + 2y\). \(y^2+2y = (y+1)^2 - 1 = \left[\sum\limits_{k=0}^{\infty} \dbinom{2k}{k}*(1/5)^k\right]^2-1=5-1=4\)
ParthKohli
  • ParthKohli
lol that was simple
anonymous
  • anonymous
I have zero supporters. R.I.P my self esteem.
ganeshie8
  • ganeshie8
If \[y = \frac{2}{5}+\frac{1\cdot 3}{2!}\left(\frac{2}{5}\right)^2 + \frac{1\cdot 3 \cdot 5}{3!}\left(\frac{2}{5}\right)^3 + \cdots\]Find the value of \(y^2 + 2y\). \(y^2+2y = (y+1)^2 - 1 = \left[\sum\limits_{k=0}^{\infty} \dbinom{2k}{k}*(1/5)^k\right]^2-1=5-1=4\) ( because maclaurin series of \(\dfrac{1}{\sqrt{1-4x}}\) is \(\sum\limits_{k=0}^{\infty} \dbinom{2k}{k}x^k\) )
anonymous
  • anonymous
If 4+6=?
ganeshie8
  • ganeshie8
If \((9+\sqrt{80})^n = I + f\) where \(I, n\) are integers and \(0
ParthKohli
  • ParthKohli
Whoa, great work! You were able to do this without knowing the general way to do these problems.
ParthKohli
  • ParthKohli
BTW it's a multi-correct question so pay attention to the other choices as well.
ganeshie8
  • ganeshie8
Oh, whats the general mehtod ? before that let me have a good look at other options..
ganeshie8
  • ganeshie8
Since I is odd, B cannot be true
ParthKohli
  • ParthKohli
Didn't you say it was even?
ganeshie8
  • ganeshie8
right hand side is even, so subtracting something less than 1 from it gives an odd integer for the integer part it was a typo earlier...
ganeshie8
  • ganeshie8
Fixed it here : If \((9+\sqrt{80})^n = I + f\) where \(I, n\) are integers and \(0
ganeshie8
  • ganeshie8
\(\color{red}{(I+f)}\color{blue}{(1-f)} = \color{red}{p^k}\color{blue}{(1-(1-q^k))} = p^kq^k= 1^k=1\) So A, C, and D ?
ParthKohli
  • ParthKohli
Yes.
AlexandervonHumboldt2
  • AlexandervonHumboldt2
IT IS A PARTY!
ikram002p
  • ikram002p
part is over alex :P
ganeshie8
  • ganeshie8
@ParthKohli may I see your solution for first q ?
ParthKohli
  • ParthKohli
Yes. We know that \(0<9 - \sqrt{80}< 1\) and consequently \(0<(9 - \sqrt{80})^n<1\). Thus, call this number \(f'\). Now consider the expansions of \(I+f\) and \(f'\).\[I+f=(9+\sqrt{80})^n = \binom{n}0 9^n + \binom{n}1 9^{n-1}\sqrt{80} + \binom{n}29^{n-2}(80) + \cdots\]\[f' = (9- \sqrt{80})^n = \binom{n}0 9^n - \binom{n}19^{n-1} \sqrt{80} + \binom{n}2 9^{n-2}(80)+\cdots \]From these expansions, we observe that \(I + f + f'\) is an integer. Thus, \(f+f'\) should also be an integer. Now \(0 < f, f' < 1\) so the only possible value of \(f+f'\) is 1. Also note that we add the two expressions \(I+f\) and \(f'\), we get \(2\times\rm something\) thus \(I + f+f' = I +1 = 2k \) which means \(I\) is an odd integer.
ganeshie8
  • ganeshie8
thats binomial theorem all the way, really clever !

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