Question

If A be the sum of the odd terms and B the sum of the even terms in the expansion of ${(x+a)}^{n}$, prove that

${A}^{2}-{B}^{2}={(x^2-a^2)}^{n}$


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Answer 1

See the reference... ........

Answer 2

ANSWER

Here we have →

$(x+a)^{n} ={ }^{n}C_{0}x^{n} + { }^{n}C_{1}x^{n-1}a^{1} + { }^{n}C_{2} x^{n-2}a^{2} ........+{ }^{n}C_{n}a^{n} --------(i)$

$(x-a)^{n} ={ }^{n}C_{0}x^{n} - { }^{n}C_{1}x^{n-1}a^{1} + { }^{n}C_{2} x^{n-2}a^{2} ........+(-1)^{n}\:{ }^{n} C_{n}a^{n} --------(ii)$

Adding (i) and (ii) we get →

$(x+a)^{n} + (x-a)^{n} =2[{ } ^{n}C_{0}x^{n} + (-1)^{n}{ }^{n} C_{n}a^{n} +....]$

$\implies \dfrac{ (x+a)^{n} + (x-a)^{n} }{2} = A[let]$

Similarly, on subtracting (ii) from (i) we get →

$\implies \dfrac{ (x+a)^{n} - (x-a)^{n} }{2} = { }^{n}C_{1}x^{n-1}a^{1} + { }^{n}C_{3}x^{n-3}a^{3} +.....$

$\implies \dfrac{ (x+a)^{n} - (x-a)^{n} }{2} =B[Let]$

Now, we know that →

(A² - B²) = (A + B)(A - B)

$\implies A^{2} - B^{2} = [ \dfrac{ (x+a)^{n} - (x-a)^{n} }{2}+ \dfrac{ (x+a)^{n} + (x-a)^{n} }{2}][ \dfrac{ (x+a)^{n} + (x-a)^{n} }{2} - \dfrac{ (x+a)^{n} - (x-a)^{n} }{2}]$

$\implies A^{2} - B^{2} = [ \dfrac{ (x+a)^{n} - (x-a)^{n} + (x+a)^{n} + (x-a)^{n} }{2}][ \dfrac{ (x+a)^{n} + (x-a)^{n} - (x+a)^{n} + (x-a)^{n} }{2}]$

$\implies A^{2} - B^{2} =(x + a)^{n} {(x - a)^{n} }$

$\implies A^{2} - B^{2} = (x^{2} - a^{2})^{n}$

HENCE PROVED.