First slide

Binomial theorem for positive integral Index

Question

If A is the sum of the odd terms and B the sum of even terms in the expansion of (x + a)ⁿ, then $A^{2} - B^{2} =$

Moderate

A

${(x^{2} + a^{2})}^{n}$
B

${(x^{2} - a^{2})}^{n}$
C

$2 {(x^{2} - a^{2})}^{n}$
D

none of these

Solution

We have,
$\begin{matrix} (x + a)^{n} =^{n} C_{6} x^{n} +^{n} C_{1} x^{n - 1} a^{1} +^{n} C_{2} x^{n - 2} a^{2} +^{n} C_{3} x^{n - 3} a^{3} + \dots +^{n} C_{n} x^{n} \\ = (^{n} C_{0} x^{n} +^{n} C_{2} x^{n - 2} a^{2} + \dots) + (^{n} C_{1} x^{n - 1} a^{1} +^{n} C_{3} x^{n - 3} a^{3} + \dots) \\ = A + B \\ (x - a)^{n} =^{n} C_{0} x^{n} -^{n} C_{1} x^{n - 1} a^{1} +^{n} C_{2} x^{n - 2} a^{2} -^{n} C_{3} x^{n - 3} a^{3} + \dots +^{n} C_{n} (- 1)^{n} a^{n} \\ = (^{n} C_{0} x^{n} +^{n} C_{2} x^{n - 2} a^{2} + \dots) - (^{n} C_{1} x^{n - 1} a^{1} +^{n} C_{3} x^{n - 3} a^{3} + \dots) \\ = A - B \\ ∴ A^{2} - B^{2} = (A + B) (A - B) = (x + a)^{n} (x - a)^{n} \\ = {(x^{2} - a^{2})}^{n} \end{matrix}$

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