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Question: If \( {\left( {7 + 4\sqrt 3 } \right)^n} = p + \beta \) , where n and p are positive integers , and ...

If (7+43)n=p+β{\left( {7 + 4\sqrt 3 } \right)^n} = p + \beta , where n and p are positive integers , and β\beta a proper fraction, show that (1−β)(p+β)=1\left( {1 - \beta } \right)\left( {p + \beta } \right) = 1 .

Explanation

Solution

Hint: To solve this question, we will use the concept of binomial theorem. According to the binomial theorem, the binomial expansion for any positive integer n is given by, (a+b)n=nC0an+nC1an−1b1+nC2an−2b2+..........+nCnbn.{\left( {a + b} \right)^n}{ = ^n}{C_0}{a^n}{ + ^n}{C_1}{a^{n - 1}}{b^1}{ + ^n}{C_2}{a^{n - 2}}{b^2} + ..........{ + ^n}{C_n}{b^n}.

Complete step-by-step answer:
Given that,
(7+43)n=p+β{\left( {7 + 4\sqrt 3 } \right)^n} = p + \beta . ………… (i)
Where n and p are positive integers and β\beta a proper fraction.
We know that the binomial expansion for any positive integer n is given by,
(a+b)n=nC0an+nC1an−1b1+nC2an−2b2+..........+nCnbn.{\left( {a + b} \right)^n}{ = ^n}{C_0}{a^n}{ + ^n}{C_1}{a^{n - 1}}{b^1}{ + ^n}{C_2}{a^{n - 2}}{b^2} + ..........{ + ^n}{C_n}{b^n}.
Expanding the equation (i) using this, we will get
(7+43)n=p+β=nC07n+nC17n−143+nC27n−2(43)2+..........+nCn(43)n.{\left( {7 + 4\sqrt 3 } \right)^n} = p + \beta { = ^n}{C_0}{7^n}{ + ^n}{C_1}{7^{n - 1}}4\sqrt 3 { + ^n}{C_2}{7^{n - 2}}{\left( {4\sqrt 3 } \right)^2} + ..........{ + ^n}{C_n}{\left( {4\sqrt 3 } \right)^n}. … (ii)
Let us take the value of (7−43)\left( {7 - 4\sqrt 3 } \right) ,
⇒0<(7−43)<1 ⇒0<(0.0718)<1  \Rightarrow 0 < \left( {7 - 4\sqrt 3 } \right) < 1 \\\ \Rightarrow 0 < \left( {0.0718} \right) < 1 \\\
(7−43)\left( {7 - 4\sqrt 3 } \right) is a fractional value which is less than 1 and greater than 0.
We will take (7−43)n{\left( {7 - 4\sqrt 3 } \right)^n} because n is a positive integer, if we take any value of n then the result will be a fraction.
Now,
(7−43)n=β′(say){\left( {7 - 4\sqrt 3 } \right)^n} = \beta '(say)
Now, we will expand this and get,
(7−43)n=β′=nC07n−nC17n−143+nC27n−2(43)2+..........+nCn(43)n.{\left( {7 - 4\sqrt 3 } \right)^n} = \beta '{ = ^n}{C_0}{7^n}{ - ^n}{C_1}{7^{n - 1}}4\sqrt 3 { + ^n}{C_2}{7^{n - 2}}{\left( {4\sqrt 3 } \right)^2} + ..........{ + ^n}{C_n}{\left( {4\sqrt 3 } \right)^n}. …….(iii)
Adding equation (ii) and (iii), we will get,
\Rightarrow {\left( {7 + 4\sqrt 3 } \right)^n} + {\left( {7 - 4\sqrt 3 } \right)^n} = p + \beta + \beta ' = 2\left\\{ {^n{C_0}{7^n}{ + ^n}{C_2}{7^{n - 2}}{{\left( {4\sqrt 3 } \right)}^2}{ + ^n}{C_4}{7^{n - 4}}{{\left( {4\sqrt 3 } \right)}^4} + .......... + .} \right\\}
All the terms that are remaining are even terms.
Thus,
\left\\{ {^n{C_0}{7^n}{ + ^n}{C_2}{7^{n - 2}}{{\left( {4\sqrt 3 } \right)}^2}{ + ^n}{C_4}{7^{n - 4}}{{\left( {4\sqrt 3 } \right)}^4} + .......... + .} \right\\} = k , where k is an even integer.
Now,
⇒(7+43)n+(7−43)n=p+β+β′=2k\Rightarrow {\left( {7 + 4\sqrt 3 } \right)^n} + {\left( {7 - 4\sqrt 3 } \right)^n} = p + \beta + \beta ' = 2k .
So, we can say that,
⇒β+β′=2k−p=int⁡eger\Rightarrow \beta + \beta ' = 2k - p = \operatorname{int} eger .
As we know that β\beta is a proper fraction, so the range of β\beta lies between 0 and 1.
i.e. 0<β<10 < \beta < 1 .
Similarly β′\beta ' is also a fraction integer, so its range will also lie between 0 and 1.
i.e. 0<β′<10 < \beta ' < 1
if we add β\beta and β′\beta ' , we will get
⇒0<β+β′<1+1 ⇒0<β+β′<2  \Rightarrow 0 < \beta + \beta ' < 1 + 1 \\\ \Rightarrow 0 < \beta + \beta ' < 2 \\\
As we can see, the value of β+β′\beta + \beta ' lies between 0 and 2.
Thus, its integer value will be,
⇒β+β′=1\Rightarrow \beta + \beta ' = 1 .
We can also write this as:
⇒β′=1−β\Rightarrow \beta ' = 1 - \beta .
According to the question, we have to show that (1−β)(p+β)=1\left( {1 - \beta } \right)\left( {p + \beta } \right) = 1 .
Taking L.H.S.
⇒(1−β)(p+β)=β′(p+β)\Rightarrow \left( {1 - \beta } \right)\left( {p + \beta } \right) = \beta '\left( {p + \beta } \right) .
Putting the values of β\beta and β′\beta ' , we get
⇒β′(p+β)=(7−43)n(7+43)n\Rightarrow \beta '\left( {p + \beta } \right) = {\left( {7 - 4\sqrt 3 } \right)^n}{\left( {7 + 4\sqrt 3 } \right)^n}.
Using the identity, anbn=(a.b)n{a^n}{b^n} = {\left( {a.b} \right)^n} , we can write this as,
⇒β′(p+β)=[(7−43)(7+43)]n\Rightarrow \beta '\left( {p + \beta } \right) = {\left[ {\left( {7 - 4\sqrt 3 } \right)\left( {7 + 4\sqrt 3 } \right)} \right]^n}.
Now, using identity, (a+b)(a−b)=a2−b2(a + b)(a - b) = {a^2} - {b^2} , we can write this as,

⇒β′(p+β)=[(72−(43)2)]n ⇒β′(p+β)=[(49−48)]n ⇒β′(p+β)=1n ⇒β′(p+β)=1  \Rightarrow \beta '\left( {p + \beta } \right) = {\left[ {\left( {{7^2} - {{\left( {4\sqrt 3 } \right)}^2}} \right)} \right]^n} \\\ \Rightarrow \beta '\left( {p + \beta } \right) = {\left[ {\left( {49 - 48} \right)} \right]^n} \\\ \Rightarrow \beta '\left( {p + \beta } \right) = {1^n} \\\ \Rightarrow \beta '\left( {p + \beta } \right) = 1 \\\

Replacing the value of β′\beta ' , we get
⇒(1−β)(p+β)=1\Rightarrow \left( {1 - \beta } \right)\left( {p + \beta } \right) = 1 .
Hence proved.

Note: Whenever we ask such questions, we have to know the method of binomial expansion. First we have to find out the given terms and then we will consider some values related to that term. Then we will expand them using binomial theorem. After that we will add their expansions and make an equation. After that we will solve that equation by considering the statements that are given in the question. Solving it step by step we will get the required answer.