Question

The integral part of $(5\sqrt{5}+11{)}^{2n+1}$ is an odd number.

• A. True
• B. False

Answer 1

The correct option is B

False

This is similar to the previous, except that it has variable power 2n +1. The power

is an odd number in this case also. We will proceed the same way we did for last

question. Before actually solving it, we can put some value for n and check the

answer. The approximate value of $5\sqrt{5}+11$ is 22.1. the integral part of it is

even(22).

$(5\sqrt{5}+11{)}^{2n+1}$ = I + f, where I is the integral part and f is the fractional part. We

want to find if I even.

Consider $(5\sqrt{5}-11{)}^{2n+1}$. It lies between 0 and 1

Or 0 < $(5\sqrt{5}-11{)}^{2n+1}$ < 1

$(5\sqrt{5}-11{)}^{2n+1}$ = f'

$(5\sqrt{5}+11{)}^{2n+1}$ = $(5\sqrt{5}{)}^{2n+1}$ + ${}^{2n+1}{C}_1$ $(5\sqrt{5}{)}^{2n}$$(11{)}^1$ + ${}^{2n+1}{C}_2$ $(5\sqrt{5}{)}^{2n-1}$$(11{)}^2$........... ..........(1)

$(5\sqrt{5}-11{)}^{2n+1}$ = $(5\sqrt{5}{)}^{2n+1}$ - ${}^{2n+1}{C}_1$ $(5\sqrt{5}{)}^{2n}$$(11{)}^1$ - ${}^{2n+1}{C}_2$ $(5\sqrt{5}{)}^{2n-1}$$(11{)}^2$........... ..........(2)

(1) - (2) gives

$(5\sqrt{5}+11{)}^{2n+1}$ - $(5\sqrt{5}-11{)}^{2n+1}$ = 2[${}^{2n+1}{C}_1$ $(5\sqrt{5}{)}^{2n}$$(11{)}^1$ + ${}^{2n+1}{C}_2$ $(5\sqrt{5}{)}^{2n-1}$$(11{)}^2$............]

We have $(5\sqrt{5}+11{)}^{2n+1}$ = I + f and $(5\sqrt{5}-11{)}^{2n+1}$ = f'

⇒ I + f - f' = 2[ ${}^{2n+1}{C}_1$ $(5\sqrt{5}{)}^{2n}$$(11{)}^1$ + ${}^{2n+1}{C}_2$ $(5\sqrt{5}{)}^{2n-1}$$(11{)}^2$........... ] .............(3)

2[ ${}^{2n+1}{C}_1$ $(5\sqrt{5}{)}^{2n}$$(11{)}^1$ + ${}^{2n+1}{C}_2$ $(5\sqrt{5}{)}^{2n-1}$$(11{)}^2$........... ] is an even integer.

Now let us consider f - f' separately,

0 < f < 1 and 0 < f' < 1

f - f' will attain its maximum value when f = 1 and f' = 0 and it will attain its

minimum value when f = 0 and f' = 1.

⇒ -1 < f - f' < 1

I + f - f' is an even integer (from(3)).

I is also an integer. For I + f - f' to be an integer, f - f' should also be an integer.

The only integer value it can take is 0, because -1 < f - f' < 1.

⇒ f - f' = 0

⇒ f = f'

⇒ I + f - f' = I = an even integer