First slide

Binomial theorem for positive integral Index

Question

Let a and b be positive integers and b not a perfect square, then for every positive integer n, the number $(a + \sqrt{b})^{n}$ is irrational. Also,
$\begin{aligned} (a + \sqrt{b})^{n} + (a - \sqrt{b})^{n} \\ = 2 [^{n} C_{0} a^{n} +^{n} C_{2} a^{n - 2} (\sqrt{b})^{2} +^{n} C_{4} a^{n - 4} (\sqrt{b})^{4} + \dots .] \end{aligned}$
Clearly, R.H.S. is an even integer say E.
Let $(a + \sqrt{b})^{n} = I + F$ where I is the integral part and
F is the fractional part of $(a + \sqrt{b})^{n} i.e., 0 < F < 1$
Let $(a - \sqrt{b})^{n} = F^{'}, 0 < F^{'} < 1$
Then, $I + F + F^{'} = E$
As, I and E are integers, F + F′ , must be an integer.
But, $0 < F + F^{'} < 2 \Rightarrow F + F^{'} = 1 \Rightarrow I = E - 1$
Thus, integral part of $(a + \sqrt{b})^{n}$ is The fractional part of $(a + \sqrt{b})^{n} is 1 - (a - \sqrt{b})^{n}$

Moderate

Question

Let $R = (5 + 2 \sqrt{6})^{n}$ and f = fractional part of R, then R (1 – f ) =

A

1
B

-1
C

0
D

none of these

Solution

Let ${(5 + 2 \sqrt{6})}^{n} = I + f$ where I and f are the integral and the fractional parts of ${(5 + 2 \sqrt{6})}^{n}$ respectively.
Let $0 < 5 - 2 \sqrt{6} < 1$ where g is a fraction.
Since $0 < 5 - 2 \sqrt{6} < 1$ , therefore $0 < {(5 - 2 \sqrt{6})}^{n} < 1$
for every positive integer n. We have
$\begin{aligned} I + f + g = {(5 + 2 \sqrt{6})}^{n} + {(5 - 2 \sqrt{6})}^{n} \\ = 2 [C_{0} 5^{n} + C_{2} (2 \sqrt{6})^{2} . {5^{n}}^{- 2} + \dots .] \end{aligned}$
$i.e., I + f + g = 2 k, k \in I^{+} - - - (1)$
$∴ f + g = 2 k - 1 = an integer ---- (2)$
Since 0 < f, g < 1, therefore we have 0 < f + g < 2 --------(3)
Thus, using (2) and (3), we have
f + g= 1
[since the only integral value in (0,2) is 1]
i.e., g =1– f .………..(4)
Therefore, we have,
$\begin{aligned} R (1 - f) = Rg = (5 + 2 \sqrt{6}) n \cdot (5 - 2 \sqrt{6}) n \\ = {(5^{2} - (2 \sqrt{6})^{2})}^{n} = 1 \end{aligned}$

Question

$[(3 + \sqrt{5})^{2 n}] + 1$ where [x] denotes the integral part of x, is divisible by

A

2^{n –1}
B

2ⁿ
C

2^{n +1}
D

none of these

Solution

Let $(3 + {\sqrt{5}}^{) 2 n} = I + f$ where I and f are the integral and the
fractional parts of $(3 + \sqrt{5})^{2 n}$ respectively.
Let ( $(3 + \sqrt{5})^{2 n} = g$ where g is a fraction.
Since 0 < 3 – 5 < 1, therefore
$0 < (3 - \sqrt{5})^{2 n} < 1$ for every positive integer n.
We have,
$\begin{aligned} I + f + g = (3 + \sqrt{5})^{2 n} + (3 - \sqrt{5})^{2 n} \\ = {(14 + 6 \sqrt{5})}^{n} + {(14 - 6 \sqrt{5})}^{n} \\ = 2 n [{(7 + 3 \sqrt{5})}^{n} + {(7 - 3 \sqrt{5})}^{n}] \\ = 2 n . 2 [C_{0} . 7 n + C_{2} (7) n^{- 2} (3 \sqrt{5})^{2} + \dots .] \\ i.e., I + f + g = (2 n^{} + 1) k, k \in I^{+} - - - - (1) \end{aligned}$
i.e., $f + g = (2 n + 1) k - I =$ an integer ……..(2)
Since 0 < f, g < 1, therefore we have
0 < f + g < 2 ……....(3)
Thus, using (2) and (3), we have
f + g = 1 ……...(4)
[since the only integral value in (0,2) is1]
Putting (4) in equation (1), we have
I + 1 = (2n ^{+ 1)}k
$∴ [(3 + \sqrt{5})^{2 n}] + 1$ is divisible by 2n + 1.

Question

If $n \in N$ such that $(7 + 4 \sqrt{3})^{n} = I + f$ ,where $I \in N$ and 0 < f < 1. Then, the value of ( I + f ) (I – f) is

A

0
B

1
C

7²ⁿ
D

2²ⁿ

Solution

. Let $g = {(7 - 4 \sqrt{3})}^{n}$ . Then, 0 < g < 1 as $0 < 7 - 4 \sqrt{3} < 1$
Now, $I + f + g = {(7 + 4 \sqrt{3})}^{n} + {(7 - 4 \sqrt{3})}^{n}$
$= 2 ({nC}_{0} 7 n + {nC}_{2} 7 n^{- 2} (4 \sqrt{3})^{2} + \dots)$
= an integer
$\Rightarrow f + g = I \Rightarrow g = I - f$
thus, $(I + f) (I - f) = (I + f) g = {(7 + 4 \sqrt{3})}^{n} {(7 - 4 \sqrt{3})}^{n} = 1$

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