Question

If P and Q are the sum of odd term and sum of even terms in the expansion of \[{{\left( x+a \right)}^{n}}\] respectively then prove that
(1) \[{{P}^{2}}-{{Q}^{2}}={{\left( {{x}^{2}}-{{a}^{2}} \right)}^{n}}\]
(2) \[4PQ={{\left( x+a \right)}^{2n}}-{{\left( x-a \right)}^{2n}}\]

Answer 1

Hint: We solve this problem by using the binomial expansion and some algebra formulas.
We have the binomial expansion of \[{{\left( x+a \right)}^{n}}\] and \[{{\left( x-a \right)}^{n}}\] as
\[{{\left( x+a \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{1}}{{x}^{n-1}}a+{}^{n}{{C}_{2}}{{x}^{n-2}}{{a}^{2}}+....+{}^{n}{{C}_{n}}{{a}^{n}}\]
\[{{\left( x-a \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}-{}^{n}{{C}_{1}}{{x}^{n-1}}a+{}^{n}{{C}_{2}}{{x}^{n-2}}{{a}^{2}}+....+{{\left( -1 \right)}^{n}}{}^{n}{{C}_{n}}{{a}^{n}}\]
We have the formula of algebra as
\[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]
\[4ab={{\left( a+b \right)}^{2}}-{{\left( a-b \right)}^{2}}\]

Complete step by step answer:
We are given that the P and Q are the sum of odd term and sum of even terms in the expansion of \[{{\left( x+a \right)}^{n}}\] respectively
We know that the binomial expansion of \[{{\left( x+a \right)}^{n}}\] as
\[{{\left( x+a \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{1}}{{x}^{n-1}}a+{}^{n}{{C}_{2}}{{x}^{n-2}}{{a}^{2}}+....+{}^{n}{{C}_{n}}{{a}^{n}}\]
By using the given condition that P and Q are sum of odd and even terms we get
\[\Rightarrow P={}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{2}}{{x}^{n-2}}{{a}^{2}}+.......\]
\[\Rightarrow Q={}^{n}{{C}_{1}}{{x}^{n-1}}a+{}^{n}{{C}_{3}}{{x}^{n-3}}{{a}^{3}}+.....\]
Now, by adding the values of P and Q we get
\[\begin{align}
  & \Rightarrow P+Q=\left( {}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{2}}{{x}^{n-2}}{{a}^{2}}+....... \right)+\left( {}^{n}{{C}_{1}}{{x}^{n-1}}a+{}^{n}{{C}_{3}}{{x}^{n-3}}{{a}^{3}}+..... \right) \\
 & \Rightarrow P+Q={}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{1}}{{x}^{n-1}}a+{}^{n}{{C}_{2}}{{x}^{n-2}}{{a}^{2}}+....+{}^{n}{{C}_{n}}{{a}^{n}} \\
 & \Rightarrow P+Q={{\left( x+a \right)}^{n}} \\
\end{align}\]
Now, by subtracting the value of Q from P then we get
\[\begin{align}
  & \Rightarrow P-Q=\left( {}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{2}}{{x}^{n-2}}{{a}^{2}}+....... \right)-\left( {}^{n}{{C}_{1}}{{x}^{n-1}}a+{}^{n}{{C}_{3}}{{x}^{n-3}}{{a}^{3}}+..... \right) \\
 & \Rightarrow P-Q={}^{n}{{C}_{0}}{{x}^{n}}-{}^{n}{{C}_{1}}{{x}^{n-1}}a+{}^{n}{{C}_{2}}{{x}^{n-2}}{{a}^{2}}+....+{{\left( -1 \right)}^{n}}{}^{n}{{C}_{n}}{{a}^{n}} \\
\end{align}\]
We know that the binomial expansion of \[{{\left( x-a \right)}^{n}}\] is given as
\[{{\left( x-a \right)}^{n}}={}^{n}{{C}_{0}}{{x}^{n}}-{}^{n}{{C}_{1}}{{x}^{n-1}}a+{}^{n}{{C}_{2}}{{x}^{n-2}}{{a}^{2}}+....+{{\left( -1 \right)}^{n}}{}^{n}{{C}_{n}}{{a}^{n}}\]
By using the above formula we get
\[\Rightarrow P-Q={{\left( x-a \right)}^{n}}\]
Now, let us prove the first result
We know that the formula of algebra as
\[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)\]
By using the above formula to P and Q we get
\[\Rightarrow {{P}^{2}}-{{Q}^{2}}=\left( P+Q \right)\left( P-Q \right)\]
By substituting the required values in above equation then we get
\[\begin{align}
  & \Rightarrow {{P}^{2}}-{{Q}^{2}}={{\left( x+a \right)}^{n}}\times {{\left( x-a \right)}^{n}} \\
 & \Rightarrow {{P}^{2}}-{{Q}^{2}}={{\left( {{x}^{2}}-{{a}^{2}} \right)}^{n}} \\
\end{align}\]
Now, let us prove the second result
We know that the formula of algebra as
\[4ab={{\left( a+b \right)}^{2}}-{{\left( a-b \right)}^{2}}\]
By using the above formula to P and Q we get
\[\Rightarrow 4PQ={{\left( P+Q \right)}^{2}}-{{\left( P-Q \right)}^{2}}\]
By substituting the required values in above equation then we get
\[\begin{align}
  & \Rightarrow 4PQ={{\left( {{\left( x+a \right)}^{n}} \right)}^{2}}-{{\left( {{\left( x-a \right)}^{n}} \right)}^{2}} \\
 & \Rightarrow 4PQ={{\left( x+a \right)}^{2n}}-{{\left( x-a \right)}^{2n}} \\
\end{align}\]
Therefore we can conclude that the required results has been proved that are
\[\therefore {{P}^{2}}-{{Q}^{2}}={{\left( {{x}^{2}}-{{a}^{2}} \right)}^{n}}\]
\[\therefore 4PQ={{\left( x+a \right)}^{2n}}-{{\left( x-a \right)}^{2n}}\]

Note:
Students may get confused between the odd terms and odd power terms.
We are given that P is the sum of odd terms of the expansion \[{{\left( x+a \right)}^{n}}\]
By using the binomial expansion we get the value of P as
\[\Rightarrow P={}^{n}{{C}_{0}}{{x}^{n}}+{}^{n}{{C}_{2}}{{x}^{n-2}}{{a}^{2}}+.......\]
But students may get confusion that P is the sum of odd power terms and take the value of P as
\[\Rightarrow P={}^{n}{{C}_{1}}{{x}^{n-1}}a+{}^{n}{{C}_{3}}{{x}^{n-3}}{{a}^{3}}+.....\]
This gives the wrong answer because the odd term refers to the number of the term.
The first term will be \[{}^{n}{{C}_{0}}{{x}^{n}}\] which is considered as an odd term.
Similarly if Q is the sum of even terms we get the value of Q as
\[\Rightarrow Q={}^{n}{{C}_{1}}{{x}^{n-1}}a+{}^{n}{{C}_{3}}{{x}^{n-3}}{{a}^{3}}+.....\]