Question

If n is positive integer, prove that the integral part of \((5\sqrt{5} + 11)^{2n+1} \) is an even number.

Answer 1

what does it mean integral part

Answer 2

1 + sqrt(3) <--- integral part is 1
2/3 + 1 <--- integral part is 1

Answer 3

okay i see

Answer 4

well u can look at the binomial expansion

Answer 5

odd^odd=odd
odd^even=odd

Answer 6

and the addition of 2 odds = even

Answer 7

therefore the integral part can only be some summation of

odd^odd and odd^even

there will be a total of 2n+2 of these addition going from 0 to 2n+1

Answer 8

therefore an even number of addition of these odd^2 and odd^even
so an even number of odd adition gives you an even number in the end

Answer 9

and a simple side note*, is that there is always a even number coming out of the choosing formula except for 0! and 1! and 2n+1 choose 2n+1, which is just 1, or u can say absence of a factor increase

Answer 10

\[
\begin{align*}
&\quad(5\sqrt{5}+11)^{2n+1}\\
&=\sum_{k=0}^{2n+1}(5\sqrt{5})^k11^{2n+1-k}\\
&=\underbrace{\sum_{k=0}^n(5\sqrt{5})^{2k}11^{2n+1-2k}}_\text{integral part}+\underbrace{\sum_{k=0}^{n}(5\sqrt{5})^{2k+1}11^{2n-2k}}_\text{non-integral part} \text{ not sure did I get this right}\\
&=11\sum_{k=0}^n125^k121^{n-k}+\sum_{k=0}^{n}(5\sqrt{5})^{2k+1}11^{2n-2k}\\
&=\underbrace{11(246)^n}_\text{even}+\sum_{k=0}^{n}(5\sqrt{5})^{2k+1}11^{2n-2k}
\end{align*}
\]

Answer 11

Oh crap I left out the binomial coefficient lol.

Answer 12

they dont matter :)

Answer 13

then @dan815 's argument has a mistake because there are exactly \(n+1\) integer terms in the expanded sum

Answer 14

Why they don't?

Answer 15

because you figured it out without using them :) once a number is even, any multiple of it is even too... so..