Question

PARTY

Answer 1

?

Answer 2

WE'RE PARTYING HERE. @GANESHIE8

Answer 3

If \((9+\sqrt{80})^n = I + f\) where \(I, n\) are integers and \(0<f<1\) then:
(A) \(I\) is an odd integer.
(B) \(I\) is an even integer.
(C) \((I+f)(1-f) = 1\)
(D) \(1-f = (9- \sqrt{80})^n\)

Answer 4

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 \(y^2 + 2y\).

Answer 5

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

Answer 6

@ParthKohli Are you on some new medication?

Answer 7

Yes.

Answer 8

I had a feeling.

Answer 9

@ParthKohli 2+2=? Solve for a cookie.

Answer 10

idk lol that's way harder than the questions I posted

Answer 11

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 \(y^2 + 2y\).

\(y^2+2y = (y+1)^2 - 1 = \left[\sum\limits_{k=0}^{\infty} \dbinom{2k}{k}*(1/5)^k\right]^2-1=5-1=4\)

Answer 12

lol that was simple

Answer 13

I have zero supporters. R.I.P my self esteem.

Answer 14

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 \(\sum\limits_{k=0}^{\infty} \dbinom{2k}{k}x^k\) )

Answer 15

If

4+6=?

Answer 16

If \((9+\sqrt{80})^n = I + f\) where \(I, n\) are integers and \(0<f<1\) then:
(A) \(I\) is an odd integer.
(B) \(I\) is an even integer.
(C) \((I+f)(1-f) = 1\)
(D) \(1-f = (9- \sqrt{80})^n\)

Notice that \(9\pm\sqrt{80}\) are the roots of quadratic eqn \(x^2-18x+1=0\).
Let \(p = 9+\sqrt{80}\) and \(q = 9-\sqrt{80}\). Since these are the roots of above eqn, we have :
\(p^2-18p+1=0 \implies p^k = 18p^{k-1}-p^{k-2}\tag{1}\)
\(q^2-18q+1=0 \implies q^k = 18q^{k-1}-q^{k-2}\tag{2}\)

Add them and get
\[p^k+q^k = 18(p^{k-1}+q^{k-1}) - (p^{k-2}+q^{k-2})\tag{3}\]

Easy to see that \(q^k \lt 1\) for all \(k\gt 0\) and by induction we can show that the right hand side of eqn (3) is even. It follows \(I\) is even.

Answer 17

Whoa, great work! You were able to do this without knowing the general way to do these problems.

Answer 18

BTW it's a multi-correct question so pay attention to the other choices as well.

Answer 19

Oh, whats the general mehtod ?
before that let me have a good look at other options..

Answer 20

Since I is odd, B cannot be true

Answer 21

Didn't you say it was even?

Answer 22

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...

Answer 23

Fixed it here :

If \((9+\sqrt{80})^n = I + f\) where \(I, n\) are integers and \(0<f<1\) then:
(A) \(I\) is an odd integer.
(B) \(I\) is an even integer.
(C) \((I+f)(1-f) = 1\)
(D) \(1-f = (9- \sqrt{80})^n\)

Notice that \(9\pm\sqrt{80}\) are the roots of quadratic eqn \(x^2-18x+1=0\).
Let \(p = 9+\sqrt{80}\) and \(q = 9-\sqrt{80}\). Since these are the roots of above eqn, we have :
\(p^2-18p+1=0 \implies p^k = 18p^{k-1}-p^{k-2}\tag{1}\)
\(q^2-18q+1=0 \implies q^k = 18q^{k-1}-q^{k-2}\tag{2}\)

Add them and get
\[p^k+q^k = 18(p^{k-1}+q^{k-1}) - (p^{k-2}+q^{k-2})\tag{3}\]

Easy to see that \(q^k \lt 1\) for all \(k\gt 0\) and by induction we can show that the right hand side of eqn (3) is even. It follows \(I\) is \(\color{red}{odd}\).

Answer 24

\(\color{red}{(I+f)}\color{blue}{(1-f)} = \color{red}{p^k}\color{blue}{(1-(1-q^k))} = p^kq^k= 1^k=1\)

So A, C, and D ?

Answer 25

Yes.

Answer 26

IT IS A PARTY!

Answer 27

part is over alex :P

Answer 28

@ParthKohli may I see your solution for first q ?

Answer 29

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.

Answer 30

thats binomial theorem all the way, really clever !